Question 14 Marks
Using factor theorem, factorize the following polynomials: $x^3 + 6x^2 + 11x + 6$
AnswerLet $x = 1$
$\text{f(1)}=1^3+6(2)^2+11(1)+6\neq0$
Let $x = -1$
$f(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6 = 12 - 12 = 0$
$\therefore$ x = -1 is a solution
$\Rightarrow x + 1 = 0$
i. e $4$ is a factor of $f(x)$
By division algorithm
$x^3 + 6x^2 + 11x + 6$
$= (x + 1)(x^2 + 5x + 6)$
$= (x + 1)(x^2 + 2x + 3x + 6)$
$= (x + 1)(x(x + 2) + 3(x + 2))$
$= (x + 1)(x + 2)(x + 3)$
$\therefore x^3 + 6x^2 + 11x + 6$
$ = (x + 1)(x + 2)(x + 3)$ View full question & answer→Question 24 Marks
Find the value of $a,$ if $x + 2$ is a factor of $4x^4 + 2x^3 - 3x^2 + 8x + 5a.$
AnswerLet $g(x) = x + 2, f(x) = 4x^4 + 2x^3 - 3x^2 + 8x + 5a.$
Let $g(x) = 0$
$\Rightarrow x + 2 = 0$
$\Rightarrow x = -2,$
$\because g(x)$ is a factor of $f(x)$
$\therefore f(-2) = 0$
$f(-2) = 4(-2)^4 + 2(-2)^3 - 3(-2)^2 + 8(-2) + 5a = 0$
$\Rightarrow 4(16) + 2(-8) - 3(4) + 8(-2) + 5a = 0$
$\Rightarrow 64 - 16 - 12 - 16 + 5a = 0$
$\Rightarrow 20 + 5a = 0$
$\Rightarrow 5a = -20$
$\Rightarrow\ \text{a}=\frac{-20}{5}=-4$
$\therefore\ \text{a}=-4$
View full question & answer→Question 34 Marks
If the polynomials $ax^3 + 3x^2 − 13$ and $2x^3 − 5x + a,$ when divided by $(x - 2)$ leave the same remainder, Find the value of $a.$
AnswerHere,
The polynomials are:
$f(x) = ax^3 + 3x^2 − 13$
$p(x) = 2x^3 − 5x + a$
equate,$ x - 2 = 0$
$x = 2$
substitute the value of $x$ in $f(x)$ and $p(x)$
$f(2) = (2)^3 + 3(2)^2 − 13$
$= 8a + 12 - 13$
$= 8a - 1 ...(1)$
$p(2) = 2(2)^3 - 5(2) + a$
$= 16 - 10 + a$
$= 6 + a ...(2)$
$f(2) = p(2)$
$\Rightarrow 8a - 1 = 6 + a$
$\Rightarrow 8a - a = 6 + 1$
$\Rightarrow 7a = 7$
$\Rightarrow a = 1$
The value of $a = 1.$
View full question & answer→Question 44 Marks
Find the rational roots of the polynomial $f(x) = 2x^3 + x^2 - 7x - 6$
AnswerGiven that $f(x) = 2x^3 + x^2 - 7x - 6$
$f(x)$ is a cubic polynomial with an integer coefficient. If the rational root in the form of $\frac{\text{p}}{\text{q}},$ the values of p are limited to factors of $6$ which are $±1, ±2, ±3, ±6$ and the values of $q$ are limited to the highest degree coefficient i.e $2$ which are $±1, ±2$
here, the possible rational roots are $±1, ±2, ±3, ±6, ±\frac{1}{2}, ±\frac{3}{2}$
Let$, x = -1$
$f(-1) = 2(-1)^3 +(-1)^{2 }- 7(-1) - 6$
$= -2 + 1 + 7 - 6$
$= -8 + 8$
$= 0$
Let,$ x = 2$
$f(-2) = 2(2)^3 + (2)^{2 }- 7(2) - 6$
$= (2 \times 8) + 4 - 14 - 6$
$= 16 + 4 -14 - 6$
$= 20 - 20$
$= 0$
Let, $\text{x}=-\frac{3}{2}$
$\text{f}=\Big(-\frac{3}{2}\Big)=2\Big(-\frac{3}{2}\Big)^3+\Big(-\frac{3}{2}\Big)^2-7\Big(-\frac{3}{2}\Big)-6$
$=2\Big(-\frac{27}{4}\Big)+\frac{9}{4}-7\Big(-\frac{3}{2}\Big)-6$
$=2\Big(-\frac{27}{4}\Big)+\frac{9}{4}-\Big(-\frac{21}{2}\Big)-6$
$=-6.75+2.25+10.5-6$
$=12.75-12.75$
$=0$
But from all the factors only $-1, 2$ and $-\frac{3}{2}$ gives the result as zero
So, the rational roots of $2x^3 + x^2 - 7x - 6$ are$ -1, 2$ and $-\frac{3}{2}$
View full question & answer→Question 54 Marks
In the following, using the remainder theorem, find the remainder when $f(x)$is divided by $g(x)$ and verify the by actual division:$ f(x) = 4x^4 - 3x^3 - 2x^2 + x - 7, g(x) = x - 1$
AnswerHere,
$f(x) = 4x^4 - 3x^3 - 2x^2 + x -7$
$g(x) = x - 1$
From, the remainder theorem when $f(x)$ is divided by $g(x) = x - (-1)$ the remainder will be equal to $f(1)$
Let$, g(x) = 0$
$\Rightarrow x - 1 = 0$
$\Rightarrow x = 1$
Substitute the value of $x$ in$ f(x)$
$f(1) = 4(1)^{4 }- 3(1)^{3 }- 2(1)^2 + 1 - 7$
$= 4 - 3 - 2 + 1 - 7$
$= 5 - 12$
$= -7$
Therefore, the remainder is $7.$
View full question & answer→Question 64 Marks
Using factor theorem, factorize the following polynomials: $x^3 + 13x^2 + 32x + 20$
AnswerLet $p(x) = x^3 + 13x^{2 }+ 32x + 20$
The factors of $20$ are $\pm1,\pm2,\pm4,\pm5\dots$
By hit and trial method
$p(-1) = (-1)^3 + 13(-1)^2 + 32(-1) + 20$
$= -1 + 13 - 32 + 20$
$= 33 - 33 = 0$
As $p(-1)$ is zero, so $x + 1$ is a factor of this polynomial $p(x).$
Let us find the quotient while dividing $x^3 + 13x^2 + 32x + 20$ by $(x + 1)$
By long division
We know that
Dividend $=$ Divisor \times Quotient $+$ Remainder
$x^3 + 13x^2 + 32x + 20$
$ = (x + 1)(x^2 + 12x + 20) + 0$
$= (x + 1)(x^2 + 10x + 2x + 20)$
$= (x + 1)[x(x + 10) + 2(x + 10)]$
$= (x + 1)(x + 10)(x + 2)$
$= (x + 1)(x + 2)(x + 10)$ View full question & answer→Question 74 Marks
In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division: $f(x) = x^3 + 4x^2 - 3x + 10, g(x) = x + 4$
AnswerHere,
$f(x) = x^3 + 4x^2 - 3x + 10$
$g(x) = x + 4$
From, the remainder theorem when $f(x)$ is divided by $g(x) = x - (-4)$ the remainder will be equal to $f(-4)$
Let,$ g(x) = 0$
$\Rightarrow x + 4 = 0$
$\Rightarrow x = -4$
Substitute the value of $x$ in $f(x)$
$f(-4) = (-4)^3 + 4(-4)^{2 }- 3(-4) + 10$
$= - 64 + (4 \times 16) + 12 + 10$
$= - 64 + 64 + 12 + 10$
$= 12 + 10$
$= 22$
Therefore, the remainder is $22.$
View full question & answer→Question 84 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:$ p(x) = x^3 - 6x^2 + 11x - 6, x = 1, 2, 3$
Answer$p(x) = x^3 - 6x^2 + 11x - 6, x = 1, 2, 3$
We know that,
$p(x) = x^3 - 6x^2 + 11x - 6$
given that the values of $x$ are $1, 2 , 3$
substitute $x = 1 in p(x)$
$p(1) = 1^{3 }- 6(1)^2 + 11(1) - 6$
$= 1 - (6 \times 1) + 11 - 6$
$= 1 - 6 + 11 - 6$
$= 0$
Now, substitute$ x = 2$ in $p(x)$
$P(2) = 2^3 - 6(2)^2 + 11(2) - 6$
$= (2 \times 3) - (6 \times 4) + (11 \times 2) - 6$
$= 8 - 24 - 22 - 6$
$= 0$
Now, substitute $x = 3$ in $p(x)$
$P(3) = 3^{3 }- 6(3)^2 + 11(3) - 6$
$= (3 \times 3) - (6 \times 9) + (11 \times 3) - 6$
$= 27 - 54 + 33 - 6$
$= 0$
Since, the result is $0$ for $x = 1, 2, 3$ these are the roots of $x^3 - 6x^2 + 11x - 6$
View full question & answer→Question 94 Marks
Using factor theorem, factorize the following polynomials: $2y^3 + y^2 - 2y - 1$
AnswerLet $p(y) = 2y^3 + y^2 - 2y - 1$
By hit and trial method
$p(1) = 2(1)^3 + (1)^2 - 2(1) - 1$
$= 2 + 1 - 2 - 1 = 0$
So,$ y -1$ is a factor of this polynomial.
Now,
By long division method,
$\therefore 2y^3 + y^2 - 2y - 1$
$ = (y - 1)(2y^2 + 3y + 1)$
$= (y - 1)(2y^2 + 2y + y + 1)$
$= (y - 1)[2y(y + 1) + 1(y + 1)]$
$= (y - 1)(y + 1)(2y + 1)$ View full question & answer→Question 104 Marks
If $\text{x}=-\frac{1}{2}$ is zero of the polynomial $p(x) = 8x^3 - ax^2 - x + 2,$Find the value of $a.$
AnswerWe know that, $p(x) = 8x^3 - ax^2 - x + 2$
Given that the value of $\text{x}=-\frac{1}{2}$
Substitute the value of $x$ in $f(x)$
$\text{p}\Big(-\frac{1}{2}\Big)=8\Big(-\frac{1}{2}\Big)^3-\text{a}\Big(-\frac{1}{2}\Big)^2-\Big(-\frac{1}{2}\Big)+2$
$=-8\Big(\frac{1}{8}\Big)-\text{a}\Big(\frac{1}{4}\Big)+\Big(\frac{1}{2}\Big)+2$
$=-1-\Big(\frac{\text{a}}{4}+\frac{1}{2}+2\Big)$
$=1-\Big(\frac{\text{a}}{4}+\frac{1}{2}\Big)$
$=\frac{3}{2}-\frac{\text{a}}{4}$
To, find the value of a, equal $\text{p}\Big(-\frac{1}{2}\Big)$ to zero
$\text{p}\Big(-\frac{1}{2}\Big)=0$
$\frac{3}{2}-\frac{\text{a}}{4}=0$
On taking $\text{L.C.M}$
$\frac{6-\text{a}}{4}=0$
$\Rightarrow6-\text{a}=0$
$\Rightarrow\text{a}=6$
View full question & answer→Question 114 Marks
In the following, using the remainder theorem, find the remainder when f(x) is divided by g(x) and verify the by actual division:
$\text{f(x)}=3\text{x}^4+2\text{x}^3-\frac{\text{x}^3}{3}-\frac{\text{x}}{9}+\frac{2}{27},$ $\text{g(x)}=\text{x}+\frac{2}{3}$
Answer Here,
$\text{f(x)}=3\text{x}^4+2\text{x}^3-\frac{\text{x}^3}{3}-\frac{\text{x}}{9}+\frac{2}{27},$
$\text{g(x)}=\text{x}+\frac{2}{3}$
From, the remainder theorem when f(x) is divided by $\text{g(x)}=\text{x}-\Big(-\frac{2}{3}\Big),$ the remainder will be equal to $\text{f}\Big(-\frac{2}{3}\Big)$
Substitute the value of x in f(x)
$\text{f}\Big(-\frac{2}{3}\Big)=3\Big(-\frac{2}{3}\Big)^4+2\Big(-\frac{2}{3}\Big)^3\\-\frac{\Big(-\frac{2}{3}\Big)^3}{3}-\Bigg[\frac{\big(-\frac{2}{3}\big)}{9}+\frac{22}{7}\Big(\frac{2}{27}\Big)\Bigg]$
$=3\Big(\frac{16}{81}\Big)+2\Big(\frac{-8}{27}\Big)-\frac{4}{(9\times3)}-\Big(\frac{-2}{(9\times3)}\Big)+\frac{2}{27}$
$=\Big(\frac{16}{27}\Big)-\Big(\frac{16}{27}\Big)-\frac{4}{27}+\Big(\frac{2}{27}\Big)+\frac{2}{27}$
$=\Big(\frac{4}{27}\Big)-\Big(\frac{4}{27}\Big)$
$=0$
Therefore, the remainder is 0.
View full question & answer→Question 124 Marks
Using factor theorem, factorize the following polynomials: $3x^3 - x^2 - 3x + 1$
AnswerLet $f(x) = 3x^3 - x^2 - 3x + 1$
The factor of the coefficient of $x^3$ is $3$. So, the possible rational roots of $f(x)$ are $\pm1$ and $\pm\frac{1}{3}$
We have,
$f(1) = 3 - 1 - 3 + 1 = 0$
$\Rightarrow (x - 1)$ is a factor of $f(x)$
$f(-1) = -3 - 1 + 3 + 1$
$\Rightarrow (x + 1)$ is a factor of $f(x)$
So, $(x - 1)$ and $(x + 1)$ are factors of $f(x)$
$\Rightarrow (x - 1)(x + 1)$ is a also a factor of $f(x)$
$\Rightarrow x^2 - 1$ is a factor of $f(x).$
Let us now divide $f(x) = 3x^3 - x^2 - 3x + 1$ by $x^2 - 1$ to get the other factors of $f(x).$
By long division, we have:
Therefore,
$3x^3 - x^2 - 3x + 1$
$ = (x^2 - 1)(3x - 1)$
Now,
$(x^2 - 1) $
$= (x - 1)(x + 1)$
Hence,
$3x^3 - x^2 - 3x + 1 $
$= (x - 1)(x + 1)(3x - 1)$ View full question & answer→Question 134 Marks
What must be added to $x^3 - 3x^2 - 12x + 19$ so that the result is exactly divisibly by $x^2 + x - 6$?
AnswerLet $p(x) = x^3 - 3x^2 - 12x + 19$ and $q(x) = x^2 + x - 6$
When $p(x)$ is divided $q(x)$ then the remainder is a linear equation, $r(x).$
Let $r(x) = ax + b$ when added to$ p(x)$ we get an expression which is divisible by $q(x).$
$f(x) = p(x) + r(x)$
$f(x) = x^3 - 3x^2 - 12x + 19 + ax + b$
$= x^3 - 3x^2 - 12x + ax + 19 + b$
$= x^3 - 3x^2 - x(12 - a) + 19 + b$
Now,
$q(x) = x^2 + x - 6 $
$= x^2 + 3x - 2x - 6 $
$= (x + 3)(x - 2)$
Also,
$f(x)$ is divisible $q(x)$
$\therefore f(-3) = 0, f(2) = 0$
$f(-3) = (-3)^3 - 3(-3)^2 - 12(-3) + 19 + a(-3) + b = 0$
$\Rightarrow -27 + 27 + 36 + 19 - 3a + b = 0$
$\Rightarrow b = 3a + 27 + 27 - 36 - 19$
$\Rightarrow b = 3a - 1 ...(1)$
$f(2) = 2^3 - 3(2)^2 - 12(2) + 19 + 2a + b = 0$
$\Rightarrow 8 - 12 - 24 + 19 + 2a + b = 0$
$\Rightarrow b = 9 - 2a ...(2)$
Equating $(1)$ and$ (2)$
$3a - 1 = 9 - 2a$
$\Rightarrow 5a = 10$
$\Rightarrow a = 2$
$\therefore b = 3a - 1 3(2) - 1 = 5 [$Substituting $b = 5$, in equation $1]$
Hence,
$ax + b = 2x + 5$
View full question & answer→Question 144 Marks
Using factor theorem, factorize the following polynomials: $x^4 - 7x^3_{ }+ 9x^2 + 7x - 10$
AnswerLet $f(x) = x^4 - 7x^3_{ }+ 9x^2 + 7x - 10$
The factors of constant term in $f(x)$ are $\pm1,\pm2,\pm5$ and $\pm10.$
We have,
$f(1) = 1 - 7 + 9 + 7 - 10 = 0$
$\Rightarrow (x - 1)$ is a factor of $f(x)$
$f(-1) = 1 + 7 + 9 - 7 - 10 = 0$
$\Rightarrow (x + 1)$ is a factor of $f(x)$
$f(2) = 16 - 56 + 36 + 14 - 10 = 0$
$\Rightarrow (x - 2)$ is a factor of $f(x)$
$f(-2) = 16 + 56 - 36 - 14 - 10 = 10$
$\Rightarrow (x + 2)$ is not a factor of $f(x)$
$f(5) = 625 - 875 + 225 + 35 - 10 = 0$
$\Rightarrow (x - 5)$ is a factor of $f(x)$
Since $f(x)$ is a polynomial of degree $4$.
So, it cannot have more than $4$ linear factors.
Thus, factors of $f(x)$ are $(x - 1), (x + 1), (x - 2)$ and $(x - 5).$
Therefore,
$f(x) = k(x - 1)(x + 1)(x - 2)(x - 5)$
$x^4 - 7x^3_{ }+ 9x^2 + 7x - 10 = k(x - 1)(x + 1)(x - 2)(x - 5) ...(1)$
Putting $x = 0$ on both sides, we get,
$-10 = k(-1)(1)(-2)(-5)$
$-10 = -10k$
$k = 1$
Substituting $k = 1$ in $(1),$ we get,
$x^4 - 7x^3_{ }+ 9x^2 + 7x - 10$
$ = (x - 1)(x + 1)(x - 2)(x - 5)$
View full question & answer→Question 154 Marks
Using factor theorem, factorize the following polynomials: $x^3 - 23x^2 + 142x - 120$
AnswerLet $p(x) = x^3 - 23x^2 + 142x - 120$
We shall now look for all the factors of -120. Some of these are $1,\pm2,\pm3,\pm,\pm4,\pm5,\pm6,\pm8$
$\pm12,\pm15,\pm20,\pm24,\pm30,\pm,60$
By trial, we find that $p(1) = 0.$
So $x - 1$ is a factor of $p(x).$
Now, we see that $x^3 - 23x^2 + 142x - 120$
$ = x^3 - x^{2 }- 22x^2 + 22x +120x - 120$
$= x^2(x - 1) - 22x(x - 1) + 120(x - 1)$
$= (x - 1)(x^2 - 22x + 120) [$Taking $(x - 1)$ common$]$
We could have also got this by dividing $p(x)$ by $x - 1.$
Now $x^2 - 22x + 120$ can be factorised either by splitting the middle term or by using the factor theorem.
By splitting the middle term, we have:
$x^2 - 22x + 120 = x^2 - 12x - 10x+ 120$
$= x(x - 12) - 10(x - 12)$
$= (x - 12)(x - 10)$
So,
$x^3 - 23x^2 + 142x - 120$
$ = (x - 1)(x - 10)(x - 12)$
View full question & answer→Question 164 Marks
Using factor theorem, factorize the following polynomials: $y^3 - 7y + 6$
AnswerLet $f(y) = y^3 - 7y + 6$
The factors of constant term in $f(y)$ are $\pm1,\pm,2,\pm3$ and $\pm6.$
We have,
$f(1) = 1 - 7 + 6 = 0$
$\Rightarrow (y - 1)$ is a factor of $f(y)$
$f(-1) = -1 + 7 + 6 = 12$
$\Rightarrow (y + 1)$ is a factor of $f(y)$
$f(2) = 8 - 14 + 6 = 0$
$\Rightarrow (y - 2)$ is a factor of $f(y)$
$f(-2) = -8 + 14 + 6 = 12$
$\Rightarrow (y + 2)$ is not a factor of $f(y)$
$f(3) = 27 - 21 + 6 = 12$
$\Rightarrow (y - 3)$ is not a factor of $f(y)$
$f(-3) = -27 + 21 + 6 = 0$
$\Rightarrow (y + 3)$ is a factor of $f(y)$
Since$ f(y)$ is a polynomial of degree $3$
So, it cannot have more than $3$ linear factors.
Thus, factors of $f(y)$ are $(y - 1)(y - 2)$ and $(y + 3).$
Therefore,
$f(y) = k(y - 1)(y - 2)(y + 3)$
$y^3 - 7y + 6 = k(y - 1)(y - 2)(y + 3) ...(1)$
Putting $y = 0$ on both sides, we get,
$6 = k(-1)(-2)(3)$
$6 = 6k$
$k = 1$
Substituting $k = 1$ in $(1),$ we get,
$y^3 - 7y + 6 $
$= (y - 1)(y - 2)(y + 3)$
View full question & answer→Question 174 Marks
Find the values of $a$ and $b,$ if $x^2 - 4$ is a factor of $ax^4 + 2x^3 - 3x^2 + bx - 4.$
AnswerLet $g(x) = x^2 - 4$ is, $f(x) = ax^4 + 2x^3 - 3x^2 + bx - 4.$
Let $g(x) = 0$
$\Rightarrow x^2 - 4 = 0$
$\Rightarrow x^2 = 4,$
$\Rightarrow \text{x}=\pm2$
Since $(x^2 - 4)$ is a factor of $f(x).$
$\therefore f(2) = 0$ and $f(-2) = 0$
$f(2) = a(2)^4 + 2(2)^3 - 3(2)^2 + b(2) - 4 = 0$
$\Rightarrow 16a + 16 - 12 + 2b - 4 = 0$
$\Rightarrow 16a + 2b = 0$
$\Rightarrow 16a = -2b$
$\Rightarrow\ \text{a}=\frac{-2\text{b}}{16}=\frac{-\text{b}}{8}\dots(1)$
Also, $f(-2) = 0$
$f(-2) = a(-2)^4 + 2(-2)^3 - 3(-2)^2 + b(-2) - 4 = 0$
$\Rightarrow 16a - 16 - 12 - 2b - 4 = 0$
$\Rightarrow 16a - 2b - 32 = 0$
$\Rightarrow 16a = 2b + 32$
$\Rightarrow\ \frac{2\text{b}+32}{16}\dots(2)$
Equating equations $(1)$ and $(2)$
$\Rightarrow\ \frac{-\text{b}}{8}=\frac{2\text{b}+32}{16}$
$\Rightarrow\ \frac{-16\text{b}}{8}=2\text{b}+32$
$\Rightarrow -2b = 2b + 32$
$\Rightarrow -4b = 32$
$\Rightarrow b = -8$
Substituting $b = -8$ in equation $(1)$
$\text{a}=\frac{-\text{b}}{8}=\frac{-(-8)}{8}=\frac{8}{8}=1$
$\therefore\text{a}=1,\text{b}=-8$
View full question & answer→Question 184 Marks
Factorize the following polynomials: $x^3 + 13x^2 + 31x - 45$ given that $x + 9$ is a factor.
AnswerLet $f(x) = x^3 + 13x^2 + 31x - 45$ be the given polynomial.
Therefore$, (x + 9)$ is a factor of the polynomial $f(x).$
Now,
$f(x) = x^2(x + 9) + 4x(x + 9) - 5(x + 9)$
$= (x + 9)(x^2 + 4x - 5)$
$= (x + 9)(x^2 +5x - x - 5)$
$= (x + 9)(x - 1)(x + 5)$
Hence $(x - 1), (x + 5)$ and $(x + 9)$ are the factors of polynomial $f(x).$
View full question & answer→Question 194 Marks
If both $x + 1$ and $x - 1$ are factors of $ax^3 + x^2 - 2x + b,$ find the values of $a$ and $b.$
AnswerLet,
$x + 1 = 0$
$x = -1$
$\because (x + 1)$ is a factor of $p(x) = ax^3 + x^2 - 2x + b$
$\therefore p(-1) = 0$
$p(-1) = a(-1)^3 + (-1)^2 - 2(-1) + b = 0$
$\Rightarrow -a + 1 + 2 + b = 0$
$\Rightarrow -a + 3 + b = 0$
$\Rightarrow a = 3 + b ...(1)$
Let,
$x - 1 = 0$
$x = 1$
$\because (x - 1)$ is a factor of $p(x)$
$\therefore p(1) = 0$
$p(1) = a(1)^3 + 1^2 -2(1) + b = 0$
$\Rightarrow a + 1 - 2 + b = 0$
$\Rightarrow a = -b + 1 ...(2)$
Equating $(1)$ and $(2)$
$\Rightarrow 3 + b = -b + 1$
$\Rightarrow b + b = 1 - 3$
$\Rightarrow 2b = -2$
$\Rightarrow b = -1$
Substituting $b = -1$ in equation$ (2)$
$a = -(-1) + 1 = 1 + 1 = 2$
$\therefore a = 2, b = -1$
View full question & answer→Question 204 Marks
Using factor theorem, factorize the following polynomials: $x^3 - 2x^2 - x + 2$
AnswerLet $f(x) = x^3 - 2x^2 - x + 2$
The factors of constant term in $f(x)$ are $\pm1,\pm2.$
We have
$f(1) = 1 - 2 - 1 + 2 = 0$
$\Rightarrow (x - 1)$ is a factor of $f(x)$
$f(-1) = -1 - 2 + 1 + 2 = 0$
$\Rightarrow (x + 1)$ is a factor of $f(x)$
$f(2) = 8 - 8 - 2 + 2 = 0$
$\Rightarrow (x - 2)$ is a factor of $f(x)$
Since $f(x)$ is a polynomial of degree $3$. So, it cannot have more than $3$ linear factors.
Thus, factors of $f(x)$ are $(x - 1), (x + 1)$ and $(x - 2).$
Therefore,
$f(x) = k(x - 1)(x + 1)(x - 2)$
$x^3 - 2x^2 - x + 2 = k(x - 1)(x + 1)(x - 2) ...(1)$
Putting $x = 0$ on both sides, we get,
$2 = k(-1)(1)(-2)$
$2 = 2k$
$k = 1$
Substituting $k = 1$ in $(1),$ we get,
$x^3 - 2x^2 - x + 2 $
$= (x - 1)(x + 1)(x - 2)$
View full question & answer→Question 214 Marks
If the polynomial $2x^3 + ax^2 + 3x - 5$ and $x^3 + x^2 - 4x + a$ leave the same remainder when divided by $x - 2,$ Find the value of $a.$
AnswerGiven, the polymials are:
$f(x) = 2x^3 + ax^2 + 3x - 5$
$p(x) = x^3 + x^2 - 4x + a$
The remainders are $f(2)$ and $p(2)$ when $f(x)$ and $p(x)$ are divided by $x - 2$
We know that,
$f(2) = p(2) ($given in problem$)$
we need to calculate $f(2)$ and $p(2)$
for,$ f(2)$
substitute $(x = 2)$ in $f(x)$
$f(2) = 2(2)^3 + a(2)^{2 }+ 3(2) - 5$
$= (2 \times 8) + 4a + 6 - 5$
$= 16 + 4a + 1$
$= 4a + 17 ...(1)$
for,$ p(2)$
substitute $(x = 2)$ in $p(x)$
$p(2) = 2^3 + 2^{2 }- 4(2) + a$
$= 8 + 4 - 8 + a$
$= 4 + a ...(2)$
Since,$ f(2) = p(2)$
Equate equation $1$ and $2$
$\Rightarrow 4a + 17 = 4 + a$
$\Rightarrow 4a - a = 4 - 17$
$\Rightarrow 3a = -13$
$\Rightarrow\ \text{a}=\frac{-13}{3}$
The value of $\text{a}=\frac{-13}{3}.$
View full question & answer→Question 224 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial $f(x)$ or, not:
$f(x) = x^3 - 6x^2 + 11x - 6, g(x) = x^2 - 3x + 2$
Answer$g(x) = x^2 - 3x + 2$
$= x^2 - 2x - x + 2$
$= x(x - 2) - 1(x - 2)$
$= (x - 2)(x - 1)$
Let $g(x) = 0$
$\Rightarrow (x - 2)(x - 1) = 0$
$If x - 2 = 0$
$ \Rightarrow x = 2$
$If x - 1 = 0$
$ \Rightarrow x = 1$
$f(2) = 2^3 - 6(2)^2 + 11(2) - 6$
$= 8 - 24 + 22 - 6$
$= 30 - 30$
$= 0$
$f(1) = 1^3 - 6(1)^2 + 11(1) - 6$
$= 1 - 6 + 11 - 6$
$= 0$
$\because f(2) = 0$ and$ f(1) = 0,$ by factor theorem,$ (x - 2)$ and $(x - 1)$ both are factors of $f(x).$
Hence, $(x - 2)(x - 1)$ is a factor of $f(x).$
View full question & answer→Question 234 Marks
In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division: $f(x) = x^4 - 3x^2 + 4, g(x) = x - 2$
AnswerHere,
$f(x) = x^4 - 3x^2 + 4$
$g(x) = x - 2$
From, the remainder theorem when $f(x)$ is divided by $g(x) = x - 2$ the remainder will be equal to $f(2)$
Let,$ g(x) = 0$
$\Rightarrow x - 2 = 0$
$\Rightarrow x = 2$
Substitute the value of $x$ in $f(x)$
$f(2) = 2^4 - 3(2)^2 + 4$
$= 16 - (3 \times 4) + 4$
$= 16 - 12 + 4$
$= 20 - 12$
$= 8$
Therefore, the remainder is $8.$
View full question & answer→Question 244 Marks
Using factor theorem, factorize the following polynomials: $y^3 - 2y^2 - 29y - 42$
AnswerLet $f(y) = y^3 - 2y^2 - 29y - 42$ be the given polynomial.
Now, putting $y = -2,$ we get
$f(-2) = (-2)^3 - 2(-2)^2 - 29(-2) - 42$
$= -8 - 8 + 58 - 42$
$= -58 + 58 = 0$
Therefore, $(y + 2)$ is a factor of polynomial $f(y)$.
Now,
$f(y) = y^2(y + 2) + 4y(y + 2) - 2(y + 2)$
$= (y + 2)(y^2 - 4y - 21)$
$= (y + 2)(y^2 - 7y + 3y - 21)$
$= (y + 2)(y + 3)(y - 7)$
Hence$ (y + 2), (y + 3)$ and $(y - 7)$ are the factors of polynomial f$(y).$
View full question & answer→Question 254 Marks
$x^4 + 10x^3 + 35x^2 + 50x + 24.$
AnswerLet $f(x) = x^4 + 10x^3 + 35x^2 + 50x + 24$
Now, putting $x = -1,$ we get
$f(-1) = (-1)^4 + 10(-1)^3 + 35(-1)^2 + 50(-1) + 24$
$= 1 - 10 + 35 - 50 + 24 = 60 - 60$
$= 0$
Therefore,$ (x + 1)$ is a factor of polynomial$ f(x).$
Now,
$f(x) = x^3(x + 1) + 9x^2(x + 1) + 26(x + 1) + 24(x + 1)$
$= (x + 1)(x^3 + 9x^2 + 26x + 24)$
$= (x + 1)g(x) ...(1)$
Where g$(x) = x^3 + 9x^2 + 26x + 24$
Putting$ x = -2,$ we get:
$g(-2) = (-2)^3 + 9(-2)^2 + 26x(-2) + 24$
$= -8 + 36 - 52 + 24 = 60 - 60$
$= 0$
Therefore, $(x + 2)$ is the factor of $g(x).$
Now,
$g(x) = x^2(x + 2) + 7x(x + 2) + 12(x + 2)$
$= (x + 2)(x^2 + 7x + 12)$
$= (x + 2)(x^2 + 4x + 3x + 12)$
$= (x + 2)(x + 3)(x + 4) ...(2)$
From equation $(1)$ and $(2),$ we get:
$f(x) = (x + 1)(x + 2)(x + 3)(x + 4)$
Hence,
$(x + 1), (x + 2), (x + 3)$ and $(x + 4)$ are the factors of polynomial $f(x).$
View full question & answer→Question 264 Marks
If $x^3 + ax^2 - bx + 10$ is divisible by $x^2 - 3x + 2,$ find the values of $a$ and $b.$
AnswerLet,
$p(x) = x^3 + ax^2 - bx + 10$
$g(x) = x^2 - 3x + 2$
$Let g(x) = 0$
$\Rightarrow x^2 - 3x + 2 = 0$
$\Rightarrow x^2 - 2x - x + 2$
$\Rightarrow x(x - 2) - 1(x - 2) =0$
$\Rightarrow x - 2 = 0, x - 1 = 0$
$\therefore x = 2, x = 1$
Since $x^2 - 3x + 2$ is a factor of $p(x)$
$\therefore (x - 2)(x - 1)$ is a factor of $p(x)$
Hence,
$p(2) = 0, p(1) = 0$
$p(2) = 2^3 + a(2)^2 - b(2) + 10 = 0$
$\Rightarrow 8 + 4a - 2b + 10 = 0$
$\Rightarrow 4a - 2b + 18 = 0$
$\Rightarrow 4a = -18 + 2b$
$\Rightarrow\ \text{a}=\frac{-18+2\text{b}}{4}=\frac{2(-9+\text{b})}{4}=\frac{-9+\text{b}}{2}\dots(1)$
$p(1) = 13 + a(1)2 - b(1) + 10 = 0$
$\Rightarrow a - b + 11 = 0$
$\Rightarrow a = b - 11 ...(2)$
Equadting $(1)$ and $(2)$
$\frac{-9+\text{b}}{2}=\text{b}-11$
$\Rightarrow -9 + b = 2b - 22$
$\Rightarrow -9 + 22 = 2b - b$
$\Rightarrow 13 = b$
Subsitituting $b = 13$ in equation $(2)$
$a = b - 11 = 13 - 11 = 2$
$\therefore a = 2, b = 13$
View full question & answer→Question 274 Marks
Using factor theorem, factorize the following polynomials: $2y^3 - 5y^2 - 19y + 42$
AnswerLet$ f(y) = 2y^3 - 5y^2 - 19y + 42$ be the given polynomial.
Now, putting $y = 2$, we get
$f(2) = 2(2)^3 - 5(2)^2 - 19(2) + 42$
$= 16 - 20 - 38 + 42$
$= -58 + 58 = 0$
Therefore, (y - 2) is a factor of polynomial $f(y).$
Now,
$f(y) = 2y^2(y - 2) - y(y - 2) - 21(y - 2)$
$= (y - 2)(2y^2 - y - 21)$
$= (y - 2)(2y^2 - 7y + 6y - 21)$
$= (y - 2)(y + 3)(2y - 7)$
Hence $(y - 2), (y + 3)$ and $(2y - 7)$ are the factors of polynomial $f(y).$
View full question & answer→Question 284 Marks
Using factor theorem, factorize the following polynomials: $x^3 - 10x^2 - 53x - 42$
AnswerLet $f(x) = x^3 - 10x^2 - 53x - 42$ be the given polynomial.
Now, putting $x = -1$, we get
$f(-1) = (-1)^3 - 10(-1)^2 - 53(-1) - 42$
$= -1 - 10 + 53 - 42$
$= -53 + 53 = 0$
Therefore, $(x + 1)$ is a factor of polynomial $f(x).$
Now,
$f(x) = x^2(x + 1) - 11x(x + 1) - 42(x + 1)$
$= (x + 1)(x^2 - 11x - 42)$
$= (x + 1)(x^2 - 14x + 3x - 42)$
$= (x + 1)(x + 3)(x - 14)$
Hence $(x + 1), (x + 3)$ and $(x - 14)$ are the factors of polynomial $f(x).$
View full question & answer→Question 294 Marks
Using factor theorem, factorize the following polynomials: $x^3 + 2x^2 - x - 2$
AnswerLet $x = 1$
$f(1) = 1^3 + 2(1)^2 - 1 - 2 = 0$
$\therefore x = 1$ is a solution
$\Rightarrow x - 1 = 0$
i. e $(x - 1)$ is a factor of $f(x)$
By division algorithm
$x^3 + 2x^2 - x - 2 = (x - 1)(x^2 + 3x + 2)$
$= (x - 1)(x^2 + 2x + x + 2)$
$= (x - 1)(x(x + 2) + 1(x + 2))$
$= (x - 1)(x + 2)(x + 1)$
$\therefore x^3 + 2x^2 - x - 2$
$ = (x - 1)(x + 2)(x + 1)$ View full question & answer→Question 304 Marks
Find the integral roots of the polynomial $f(x) = x^{3 }+ 6x^2 + 11x + 6$
AnswerGiven, that $f(x) = x^3 + 6x^2 + 11x + 6$
Clearly we can say that, the polynomial $f(x)$ with an integer coefficient and the highest degree term coefficient which is known as leading factor is 1.
So, the roots of $f(x)$ are limited to integer factor of $6$, they are $±1, ±2, ±3, ±6$
Let $x = -1$
$f(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6$
$= -1 + 6 - 11 + 6$
$= 0$
Let $x = -2$
$f(-2) = (-2)^3 + 6(-2)^2 + 11(-2) + 6$
$= -8 - (6 \times 4) - 22 + 6$
$= -8 + 24 - 22 + 6$
$= 0$
Let $x = -3$
$f(-3) = (-3)^3 + 6(-3)^2 + 11(-3) + 6$
$= -27 - (6 \times 9) - 33 + 6$
$= –27 + 54 - 33 + 6$
$= 0$
But from all the given factors only $-1, -2, -3$ gives the result as zero.
So$,$ the integral multiples of $x^3 + 6x^2 + 11x + 6$ are $-1, -2, -3.$
View full question & answer→Question 314 Marks
Show that $(x - 2), (x + 3)$ and $(x - 4)$ are factors of $x^3 - 3x^2 - 10x + 24.$
Answer$F(x) = x^3 - 3x^2 - 10x + 24$
Let,
$x - 2 = 0$
$\Rightarrow x = 2$
$f(2) = 2^3 - 3(2)^2 - 10(2) + 24$
$= 8 - 12 - 20 + 24$
$= 32 - 32 = 0$
Let,
$x + 3 = 0$
$\Rightarrow x = -3$
$f(-3) = (-3)^3 - 3(-3)^2 - 10(-3) + 24$
$= -27 - 3(9) + 30 + 24$
$= -27 - 27 + 30 + 24$
$= -54 + 54 = 0$
Let,
x - 4 = 0
$\Rightarrow x = 4$
$f(4) = 4^3 - 3(4)^2 - 10(4) + 24$
$= 64 - 3(16) - 40 + 24$
$= 64 - 48 - 40 + 24$
$= 88 - 88 = 0$
$\because f(2) = 0, f(-3) = 0, f(4) = 0$
$\therefore$ By factore theoram $(x - 2), (x + 3)$ and $(x - 4)$ are factors of $x^3 - 3x^2 - 10x + 24.$
View full question & answer→Question 324 Marks
$x^4 - 2x^3 - 7x^2 + 8x + 12.$
AnswerLet $f(x) = x^4 - 2x^3 - 7x^2 + 8x + 12$
The factors of constant term in $f(x)$ are $\pm1,\pm2,\pm3,\pm4,\pm6$ and $\pm12.$
We have,
$f(1) = 1 - 2 - 7 + 8 + 12 = 12$
$\Rightarrow (x - 1)$ is not a factor of $f(x)$
$f(-1) = 1 + 2 - 7 - 8 + 12 = 0$
$\Rightarrow (x + 1)$ is a factor of $f(x)$
$f(2) = 16 - 16 - 28 - 16 + 12 = 0$
$\Rightarrow (x - 2)$ is a factor of $f(x)$
f($-2) = 16 + 16 - 28 - 16 + 12 = 0$
$\Rightarrow (x + 2)$ is a factor of $f(x)$
f($3) = 81 - 54 - 63 + 24 + 12 = 0$
$\Rightarrow (x - 3)$ is a factor of $f(x)$
Since $f(x)$ is a polynomial of degree $4.$
So, it cannot have more than 4 linear factors.
Thus, factors of $f(x)$ are $(x +1), (x - 2), (x + 2) $and $(x - 3).$
Therefore,
$f(x) = k(x + 1)(x + 2)(x - 2)(x - 3)$
$x^4 - 2x^3 - 7x^2 + 8x + 12 $
$= k(x + 1)(x + 2)(x - 2)(x - 3) ...(1)$
Putting $x = 0$ on both sides, we get,
$12 = k(1)(2)(-2)(-3)$
$12 = 12k$
$k = 1$
Substituting $k = 1$ in $(1),$ we get,
$x^4 - 2x^3 - 7x^2 + 8x + 12 $
$= k(x + 1)(x + 2)(x - 2)(x - 3)$
View full question & answer→Question 334 Marks
If $x = 0$ and $x = -1$ are the roots of the polynomial $f(x) = 2x^{3 }- 3x^2 + ax + b,$ Find the of $a$ and $b$.
AnswerWe know that, $f(x) = 2x^{3 }- 3x^2 + ax + b$
Given, the values of x are $0$ and $-1$
Substitute $x = 0$ in $f(x)$
$f(0) = 2(0)^{3 }- 3(0)^2 + a(0) + b$
$= 0 - 0 + 0 + b$
$= b ...(1)$
Substitute $x = (-1)$ in $f(x)$
$f(-1) = 2(-1)^{3 }- 3(-1)^2 + a(-1) + b$
$= -2 - 3 - a + b$
$= -5 - a + b ...(2)$
We need to equate equations $1$ and $2$ to zero
$b = 0$ and $-5 - a + b = 0$
since, the value of $b$ is zero
substitute $b = 0$ in equation $2$
$\Rightarrow -5 - a = -b$
$\Rightarrow -5 - a = 0$
$a = -5$
the values of a and b are $-5$ and $0$ respectively.
View full question & answer→Question 344 Marks
Using factor theorem, factorize the following polynomials: $x^3 - 6x^2 + 3x + 10$
AnswerLet $x = 2$
$f(2) = 2^3 + 6(2)^2 + 3(2) + 10$
$ = 8 - 24 + 6 + 10 = 0$
$\therefore x = 2 $ is a solution $f(x)$
$i. e (x - 2)$ is a factor of $f(x)$
By division algorithm
$x^3 - 6x^2 + 3x + 10$
$= (x - 2)(x^2 - 4x - 5)$
$= (x - 2)(x^2 - 5x + x - 5)$
$= (x - 2)(x(x - 5) + 1(x - 5))$
$= (x - 2)(x - 5)(x + 1)$
$\therefore x^3 - 6x^2 + 3x + 10$
$ = (x - 2)(x - 5)(x + 1)$ View full question & answer→Question 354 Marks
Using factor theorem, factorize the following polynomials: $x^3 - 3x^2 - 9x - 5$
AnswerLet $p(x) = x^3 - 3x^2 - 9x - 5$
The factors of $5$ are $\pm1,\pm5.$
By hit and trial method
$p(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5$
$= -1 - 3 + 9 - 5 = 0$
So $x + 1$ is a factor of this polynomial $p(x).$
Let us find the quotient while dividing $x^3 - 3x^2 - 9x - 5 by x + 1$
By long division
Now,
Dividend $=$ Divisor $\times$ Quotient $+$ Remainder
$\therefore x^3 - 3x^2 - 9x - 5$
$= (x + 1)(x^2 - 4x - 5) + 0$
$= (x + 1)(x^2 - 5x + x - 5)$
$= (x + 1)[x(x - 5) + 1(x - 5)]$
$= (x + 1)(x - 5)(x + 1)$
$= (x - 5)(x + 1)(x + 1)$ View full question & answer→Question 364 Marks
$2x^4 - 7x^3 - 13x^2 + 63x - 45$
AnswerLet $f(x) = 2x^4 - 7x^3 - 13x^2 + 63x - 45$ be the given polynomial.
Now, putting $x = 1,$ we get
$f(1) = 2(1)^4 - 7(1)^3 - 13(1)^2 + 63(1) - 45$
$= 2 - 7 - 13 + 63 - 45 = 65 - 65$
$= 0$
Therefore,$ (x - 1)$ is a factor of polynomial $f(x).$
Now,
$f(x) = 2x^3(x - 1) - 5x^2(x - 1) - 18x(x - 1) + 45(x - 1)$
$= (x - 1)(2x^3 - 5x^2 - 18x + 45)$
$= (x - 1)g(x) ...(1)$
Where $g(x) = 2x^3 - 5x^2 - 18x + 45$
Putting $x = 3,$ we get:
$g(3) = 2(3)^3 - 5(3)^2 - 18(3) + 45$
$= 54 - 45 - 54 + 45$
$= 0$
Therefore,$ (x - 3)$ is the factor of $g(x).$
Now,
$g(x) = 2x^2(x - 3) + x(x - 3) - 15(x - 3)$
$= (x - 3)(2x^2 + x - 15)$
$= (x - 3)(2x^2 + 6x - 5x - 15)$
$= (x - 3)(x + 3)(2x - 5) ...(2)$
From equation $(1)$ and $(2),$ we get:
$f(x) = (x - 1)(x - 3)(x + 3)(2x - 5)$
Hence,
$(x - 1), (x - 3), (x + 3)$ and $(2x - 5)$ are the factors of polynomial $f(x).$
View full question & answer→Question 374 Marks
In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division: $f(x) = 2x^4 - 6x^3 + 2x^2 - x + 2, g(x) = x + 2$
AnswerHere,
$f(x) = 2x^4 - 6x^3 + 2x^2 - x + 2$
$g(x) = x + 2$
From, the remainder theorem when $f(x)$ is divided by $g(x) = x - (-2)$ the remainder will be equal to 4
Let,$ g(x) = 0$
$\Rightarrow x + 2 = 0$
$\Rightarrow x = -2$
Substitute the value of $x$ in $f(x)$
$f(-2) = 2(-2)^4 - 6(-2)^3 + 2(-2)^2 - (-2) + 2$
$= (2 \times 16) - (6 \times (-8)) + (2 \times 4) + 2 + 2$
$= 32 + 48 + 8 + 2 + 2$
$= 92$
Therefore, the remainder is $92.$
View full question & answer→Question 384 Marks
In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division: $f(x) = 9x^3 - 3x^2 + x - 5, \text{g(x)}=\text{x}-\frac{2}{3}$
AnswerHere,$f(x) = 9x^3 - 3x^2 + x - 5,$
$\text{g(x)}=\text{x}-\frac{2}{3}$
From, the remainder theorem when $f(x)$ is divided by $\text{g(x)}=\text{x}-\frac{2}{3}$ the remainder will be equal to $\text{f}\Big(\frac{2}{3}\Big)$
Let, $g(x) = 0$
$\Rightarrow\ \text{x}-\frac{2}{3}=0$
$\Rightarrow\ \text{x}=\frac{2}{3}$
Substitute the value of $x$ in $f(x)$
$\text{f}\Big(\frac{2}{3}\Big)=9\Big(\frac{2}{3}\Big)-3\Big(\frac{2}{3}\Big)^2+\Big(\frac{2}{3}\Big)-5$
$=9\Big(\frac{8}{27}\Big)-3\Big(\frac{4}{9}\Big)+\frac{2}{3}-5$
$=\Big(\frac{8}{3}\Big)-\Big(\frac{4}{3}\Big)+\frac{2}{3}-5$
$=\frac{8-4+2-15}{3}$
$=\frac{10-19}{3}$
$=\frac{-9}{3}$
$=-3$
Therefore, the remainder is $-3.$
View full question & answer→