Question 13 Marks
Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by:$5 + 2x$
AnswerHere,
$f(x) = x^3 + 3x^2 + 3x + 1$
By remainder theorem
$5 + 2x = 0$
$2x = -5$
$\text{x}=\frac{-5}{2}$
substitute the value of $x$ in $f(x)$
$\text{f}\Big(\frac{-5}{2}\Big)=\Big(\frac{-5}{2}\Big)^3+3\Big(\frac{-5}{2}\Big)+1$
$=\frac{-125}{8}+3\Big(\frac{25}{4}\Big)+3\Big(\frac{-5}{2}\Big)+1$
$=\frac{-125}{8}+\frac{75}{4}-\frac{15}{2}+1$
Taking $\text{L.C.M}$
$=\frac{-125+150-50+8}{8}$
$=\frac{-27}{8}$
View full question & answer→Question 23 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial $f(x)$ or, not:
$f(x) = 2x^3 - 9x^2 + x + 12, g(x) = 3 - 2x$
AnswerLet $g(x) = 0$
$\Rightarrow 3 - 2x = 0$
$\Rightarrow 3 = 2x$
$\Rightarrow\ \text{x}=\frac{2}{3}$
$\text{f}\Big(\frac{3}{2}\Big)=2\Big(\frac{3}{2}\Big)^3+\Big(\frac{3}{2}\Big)^2-\Big(\frac{3}{2}\Big)+12$
$=2\Big(\frac{27}{8}\Big)+9\Big(\frac{9}{4}\Big)-\Big(\frac{3}{2}\Big)+12$
$=\frac{27}{4}-\frac{81}{4}+\frac{3}{2}+12$
$=\frac{27-81+6+48}{4}$
$=\frac{81-81}{4}$
$=\frac{0}{4}$
$=0$
$\because\ \text{f}\Big(\frac{3}{2}\Big)=0,$ by factor theorem, $3 - 2x$ is a factor of $f(x).$
View full question & answer→Question 33 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{f(x)}=2\text{(x)}+1,\text{x}=\frac{1}{2}$
Answer$\text{f(x)}=2\text{(x)}+1,\text{x}=\frac{1}{2}$
We know that,
$\text{f(x)}=2\text{(x)}+1$
Given that $\text{x}=\frac{1}{2}$
Substitute the value of x and f(x)
$\text{f}\Big(\frac{1}{2}\Big)=2\Big(\frac{1}{2}\Big)+1$
$=1+1$
$=2\neq0$
Since, the result is not equal to zero
$\text{x}=\frac{1}{2}$ is the root of 2x + 1
View full question & answer→Question 43 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case: $f(x) = x^2 - 1, x = 1, -1$
Answer$f(x) = x^2 - 1, x = (1, -1)$
we know that,
$f(x) = x^2 - 1$
Given that$ x = (1, -1)$
substitute$ x = 1$ in $f(x)$
$f(1) = 1^2 - 1$
$= 1 - 1$
$= 0$
Now, substitute $x = (-1)$ in $f(x)$
$f(-1) = (-1)^2 - 1$
$= 1 - 1$
$= 0$
Since, the results when $x = (1, -1)$ are $0$ they are the roots of the polynomial $f(x) = x^2 - 1$
View full question & answer→Question 53 Marks
Find the value of a such that $(x - 4)$ is a factors of $5x^3 - 7x^2 - ax - 28.$
AnswerLet $g(x) = x - 4, f(x) = 5x^3- 7x^2 - ax - 28$
Let $g(x) = 0$
$\Rightarrow x - 4 = 0$
$\Rightarrow x = 4,$
Since $(x - 4)$ is a factor of $f(x).$
$\therefore f(4) = 0$
$f(4) = 5(4)^3 - 7(4)^2 - a(4) - 28 = 0$
$\Rightarrow 5(64) - 7(16) - 4a - 28 = 0$
$\Rightarrow 320 - 112 - 4a - 28 = 0$
$\Rightarrow 180 - 4a = 0$
$\Rightarrow 4a = 180$
$\Rightarrow\ \text{a}=\frac{180}{4}$
$=45$
View full question & answer→Question 63 Marks
Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by: $\text{x}+\pi$
AnswerHere,
$f(x) = x^3 + 3x^2 + 3x + 1$
By remainder theorem
$\text{x}+\pi=0$
$\Rightarrow\ \text{x}=-\pi$
Substitute the value of $x$ in $f(x)$
$\text{f}(-\pi)=(-\pi)^3+3(-\pi)^2+3(-\pi)+1$
$=(-\pi)^3+3(-\pi)^2+3(-\pi)+1$
View full question & answer→Question 73 Marks
Factorize the following polynomials: $4x^3 + 20x^2 + 33x + 18$ given that $2x + 3$ is a factor.
AnswerLet $f(x) = 4x^3 + 20x^2 + 33x + 18$ be the given polynomial.
Therefore$, (2x + 3)$ is a factor of the polynomial $f(x).$
Now,
$f(x) = 2x^2(2x + 3) + 7x(2x + 3) + 6(2x + 3)$
$= (2x + 3)(2x^2 + 7x + 6)$
$= (2x + 3)(2x^2 +4x + 3x + 6)$
$= (2x + 3)(2x + 3)(x + 2)$
Hence $(x + 2), (2x + 3)$ and $(2x + 3)$ are the factors of polynomial $f(x).$
View full question & answer→Question 83 Marks
For what value of a is $(x - 5)$ a factor of $x^3{ }- 3x^2 + ax - 10$ ?
AnswerLet,
$x - 5 = 0$
$x = 5$
$\because (x - 5)$ is a factor of$ x^3 - 3x^2 + ax - 10$
$\therefore f(5) = 0$
$f(5) = 5^3 - 3(5)^2 + a(5) - 10 = 0$
$\Rightarrow 125 - 3 \times 25 + 5a - 10 = 0$
$\Rightarrow 125 - 85 + 5a = 0$
$\Rightarrow 40 + 5a = 0$
$\Rightarrow 5a = -40$
$\Rightarrow\ \text{a}=\frac{-40}{5}=-8$
View full question & answer→Question 93 Marks
What must be added to $3x^3 + x^2 - 22x + 9$ so that the result is exactly divisible by $3x^2 + 7x - 6$?
AnswerWe know that, Dividend $=$ Divisor $\times$ Quotient $+$ Remainder
Dividend $= 3x^3 + x^2 - 22x + 9$
Divisor $= 3x^2 + 7x - 6$
Remainder $= -2x - 3$
So$, -(-2x - 3) = 2x + 3$ should be added to $3x^3 + x^2 - 22x + 9$ to make it exactly divisible by $3x^2 + 7x - 6.$ View full question & answer→Question 103 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: $f(x) = 4x^3 - 12x^2 + 14x - 3, g(x) = 2x - 1$
AnswerHere,
$f(x) = 4x^3 - 12x^2 + 14x - 3$
$g(x) = 2x - 1$
From, the remainder theorem when $f(x)$ is divided by $\text{g(x)}=\Big(\text{x}-\frac{1}{2}\Big),$ the remainder will be equal to $\text{f}\Big(\frac{1}{2}\Big).$
Let, $g(x) = 0$
$\Rightarrow 2x - 1 = 0$
$\Rightarrow 2x = 1$
$\Rightarrow\ \text{x}=\frac{1}{2}$
Substitute the value of $x$ in $f(x)$
$\text{f}\Big(\frac{1}{2}\Big)=4\Big(\frac{1}{2}\Big)^3-12\Big(\frac{1}{2}\Big)^2+14\Big(\frac{1}{2}\Big)-3$
$=4\Big(\frac{1}{8}\Big)-12\Big(\frac{1}{4}\Big)+4\Big(\frac{1}{2}\Big)-3$
$=\Big(\frac{1}{2}\Big)-3+7-3$
$=\Big(\frac{1}{2}\Big)+1$
Taking $\text{L.C.M.}$
$=\Big(\frac{2+1}{2}\Big)$
$=\Big(\frac{3}{2}\Big)$
Therefore, the remainder is $\Big(\frac{3}{2}\Big).$
View full question & answer→Question 113 Marks
If $x = 2$ is a root of the polynomial $f(x) = 2x^2 - 3x + 7a,$ Find the value of $a.$
AnswerWe know that, $f(x) = 2x^2 - 3x + 7a$
Given that $x = 2$ is the root of $f(x)$
Substitute the value of $x$ in $f(x)$
$f(2) = 2(2)^{2 }- 3(2) + 7a$
$= (2 \times 4) - 6 + 7a$
$= 8 - 6 + 7a$
$= 7a + 2$
Now, equal $7a + 2$ to zero
$⟹ 7a + 2 = 0$
$⟹ 7a = -2$
$⟹ a = -27$
The value of $\text{a}=-\frac{2}{7}$
View full question & answer→Question 123 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial $f(x)$ or, not:
$f(x) = 3x^3 + x^2 - 20x +12, g(x) = 3x - 2$
AnswerLet $g(x) = 0$
$\Rightarrow 3x - 2 = 0$
$\Rightarrow\ \text{x}=\frac{2}{3}$
$\text{f}\Big(\frac{2}{3}\Big)=3\Big(\frac{2}{3}\Big)^3+\Big(\frac{2}{3}\Big)^2-20\Big(\frac{2}{3}\Big)+12$
$=\frac{24}{27}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{24+12-360+324}{27}$
$=\frac{360-360}{27}$
$=\frac{0}{27}$
$=0$
$\because\ \text{f}\Big(\frac{2}{3}\Big)=0,$ by factor theorem$, 3x - 2$ is a factor of $f(x).$
View full question & answer→Question 133 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case: $\text{g(x)}=3\text{x}^2-2,\text{x}=\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$
Answer$\text{g(x)}=3\text{x}^2-2,\text{x}=\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}$
We know that
$\text{g(x)}=3\text{x}^2-2$
Given that,
$\text{x}=\Big(\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}\Big)$
Substitute $\text{x}=\frac{2}{\sqrt{3}}$ in g(x)
$\text{g}\Big(\frac{2}{\sqrt{3}}\Big)=3\Big(\frac{2}{\sqrt{3}}\Big)^2-2$
$=3\Big(\frac{4}{3}\Big)-2$
$=4-2$
$=2\neq0$
Now, Substitute $\text{x}=-\frac{2}{\sqrt{3}}$ in g(x)
$\text{g}\Big(\frac{-2}{\sqrt{3}}\Big)=3\Big(\frac{-2}{\sqrt{3}}\Big)^2-2$
$=3\Big(\frac{4}{3}\Big)-2$
$=4-2$
$=2\neq0$
Since, the results when
$\text{x}=\Big(\frac{2}{\sqrt{3}},\frac{-2}{\sqrt{3}}\Big)$ are not $0,$ they are roots of $3x^2 - 2$
View full question & answer→Question 143 Marks
$If x - 2$ is a factor of the following two polynomials, find the values of a in case: $x^5 - 3x^4 - ax^3 + 3ax^2 + 2ax + 4$
Answer$x^5 − 3x^4 − ax^3 + 3ax^2 + 2ax + 4$
Let,
$x - 2 = 0$
$\therefore x = 2$
$\because x - 2$ is a factor of $p(x) = x^5 − 3x^4 − ax^3 + 3ax^2 + 2ax + 4$
$\therefore p(2) = 0$
$p(2) = 2^5 - 3(2)4 - a(2)^3 + 3 \times a \times 2^2 + 2 \times 2 \times a + 4 = 0$
$\Rightarrow 32 - 48 - 8a + 12a + 4a + 4 = 0$
$\Rightarrow 8a - 12 = 0$
$\Rightarrow 8a = 12$
$\Rightarrow\ \text{a}=\frac{12}{8}=\frac{3}{2}$
View full question & answer→Question 153 Marks
If $x - 2$ is a factor of the following two polynomials, find the values of a in case: $ x^3 - 2ax^2 + ax - 1$
Answer$x^3 - 2ax^2 + ax - 1$
Let,
$x - 2 = 0$
$\therefore x = 2$
$\because x - 2$ is a factor of $p(x) = x^3 - 2ax^2 + ax - 1$
$\therefore p(2) = 0$
$p(2) = 2^3 - 2a(2)^2 + 2a - 1 = 0$
$\Rightarrow 8 - 8a + 2a - 1 = 0$
$\Rightarrow 7 - 6a = 0$
$\Rightarrow 6a = 7$
$\Rightarrow\ \text{a}=\frac{7}{6}$
View full question & answer→Question 163 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{f(x)}=5\text{x}-\pi,\text{x}=\frac{4}{5}$
Answer$\text{f(x)}=5\text{x}-\pi,\text{x}=\frac{4}{5}$
we know that,
$\text{f(x)}=5\text{x}-\pi$
Given that, $\text{x}=\frac{4}{5}$
Substitute the value of x in f(x)
$\text{f}\Big(\frac{4}{5}\Big)=5\Big(\frac{4}{5}\Big)-\pi$
$=4-\pi$
$\neq0$
Since, the result is not equal to zero, $\text{x}=\frac{4}{5}$ is not the root of the polynomial $5\text{x}-\pi$
View full question & answer→Question 173 Marks
Find the value $k$ if $x - 3$ is a factor of $k^2x^3 - kx^2 + 3kx - k.$
AnswerLet $g(x) = 0$
$\Rightarrow x - 3 = 0$
$\Rightarrow x = 3,$
Since $(x - 3)$ is a factor of $f(x)$
$\therefore f(3) = 0$
$f(3) = k^23^3 - k3^2 + 3k(3) - k = 0$
$\Rightarrow 27k^2 - 9k + 9k - k = 0$
$\Rightarrow 27k^2 - k = 0$
$\Rightarrow k(27k - 1) = 0$
$\therefore k = 0, 27k - 1 = 0$
$27k = 1$
$\text{k}=\frac{1}{27}$
Hence $k = 0, \text{k}=\frac{1}{27}$
View full question & answer→