Question 12 Marks
What must be subtracted from $x^3 - 6x^2 - 15x + 80$ so that the result is exactly divisible by $x^2 + x - 12$?
Answer
$\therefore 4x - 4$ should subtracted from $x^3 - 6x^2- 15x + 80$
So that the result is exactly divisible by $x^2 + x - 12.$ View full question & answer→Question 22 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case: $f(x) = x^2, x = 0$
Answer$f(x) = x^2, x = 0$
we know that $, f(x) = x^2$
Given that value of x is $'0\ '$
Substitute the value of $x$ in $f(x)$
$f(0) = 0^2$
$= 0$
Since, the result is zero, $x = 0$ is the root of $x^2$
View full question & answer→Question 32 Marks
Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by: $x$
AnswerHere,
$f(x) = x^3 + 3x^2 + 3x + 1$
By remainder theorem
$x = 0$
substitute the value of $x$ in $f(x)$
$f(0) = 0^{3 }+ 3(0)^2 + 3(0) + 1$
$= 0 + 0 + 0 + 1$
$= 1$
View full question & answer→Question 42 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial $f(x)$ or, not:
$f(x) = 3x^4 + 17x^3 + 9x^2 - 7x - 10; g(x) = x + 5$
AnswerLet $g(x) = 0$
$\Rightarrow x + 5 = 0$
$\Rightarrow x = -5$
Now,
$f(-5) = 3(-5)^4 + 17(-5)^3 + 9(-5)^2 - 7(-5) - 10$
$= 3(625) + 17(-125) + 9(25) + 35 - 10$
$= 1875 - 2125 + 225 + 35 - 10$
$\because f(-5) = 0,$ by factor theorem $x + 5$ is a factor of $f(x).$
View full question & answer→Question 52 Marks
If $x + 1$ is a factor of $x^3 + a,$ then write the value of $a$.
AnswerAs $(x + 1)$ is a factor of polynomial $f(x) = x^3 + a.$
$i. e. f(-1) = 0$
$(-1)^3+ a = 0$
$\Rightarrow a = 1$
Thus, the value of $a = 1.$
View full question & answer→Question 62 Marks
In the following two polynomials, find the value of a, if $x + a$ is a factor. $x^3 + ax^2 - 2x + a + 4$
Answer$x^3 + ax^2 - 2x + a + 4$
Let,
$x + a = 0$
$\Rightarrow x = -a$
$\because (x + a) $ is a factor of $f(x) = x^3 + ax^2 - 2x + a + 4$
$\therefore p(-a) = 0$
$p(-a)$
$\Rightarrow (-a)^3 + (-a)^2 - 2(-a) + a + 4 = 0$
$\Rightarrow -a^3 + a^3 + 2a + a + 4 = 0$
$\Rightarrow 3a + 4 = 0$
$\Rightarrow 3a = -4$
$\Rightarrow\ \text{a}=\frac{-4}{3}$
View full question & answer→Question 72 Marks
If $\text{x}=\frac{1}{2}$ is a zero of the polynomial $f(x) = 8x^3 + ax^2 - 4x + 2,$ find the value of $a.$
AnswerSince $\text{x}=\frac{1}{2}$ is a zero of polynomial $f(x).$
Therefore $\text{f}\Big(\frac{1}{2}\Big)=0$
$\Rightarrow\ 8\Big(\frac{1}{2}\Big)^3+\text{a}\Big(\frac{1}{2}\Big)^2-4\Big(\frac{1}{2}\Big)+2=0$
$\Rightarrow\ 1+\frac{\text{a}}{4}-2+2=0$
$\Rightarrow\ \text{a}=-4$
The value of a is $-4.$
View full question & answer→Question 82 Marks
If $f(x) = 2x^3 - 13x^2 + 17x + 12,$ Find: $f(-3)$
AnswerThe given polynomial is $f(x) = 2x^3 - 13x^2 + 17x + 12$
$f(-3)$
we need to substitute the $(-3)$ in $f(x)$
$f(-3) = 2(-3)^3 - 13(-3)^2 + 17(-3) + 12$
$= (2 \times (- 27)) - (13 \times 9) - ( 17 \times 3) + 12$
$= -54 - 117 - 51 + 12$
$= -210$
therefore $f(-3) = -210$
View full question & answer→Question 92 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial $f(x)$ or, not:
$f(x) = x^3 - 6x^2 - 19x + 84, g(x) = x - 7$
AnswerLet $g(x) = 0$
$\Rightarrow x - 7 = 0$
$\Rightarrow x = 7$
$f(7) = 7^3 - 6(7)^2- 19(7) + 84$
$= 343 - 6(49) - 19(7) + 84$
$= 343 - 294 - 133 + 84$
$= 427 - 427 = 0$
$\because f(7) = 0,$ by factor theorem $x - 7$ is a factor of $f(x).$
View full question & answer→Question 102 Marks
Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by: $x + 1$
AnswerHere,$ f(x) = x^3 + 3x^2 + 3x + 1$
By remainder theorem
$x + 1 = 0$
$\Rightarrow x = -1$
substitute the value of $x$ in $f(x)$
$f(-1) = (−1)^3 + 3(−1)^2 + 3(−1) + 1$
$= -1 + 3 - 3 + 1$
$= 0$
View full question & answer→Question 112 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial $f(x)$ or, not:
$f(x) = x^5 + 3x^4 - x^3 - 3x^2 + 5x + 15, g(x) = x + 3$
AnswerLet $g(x) = 0$
$\Rightarrow x + 3 = 0$
$\Rightarrow x = -3$
$f(-3) = (-3)^5 - 3(-3)^4 -(-3)^3 - 3(-3)^2 + 5(-3) + 15$
$= -243 + 243 + 27 - 27 - 15 + 15 $
$= 0$
$\because f(-3) = 0,$ by factor theorem $x + 3$ is a factor of $f(x).$
View full question & answer→Question 122 Marks
If $f(x) = x^4 - 2x^3 + 3x^2 - ax - b$ when divided $by x - 1,$ the remainder is $6,$ then find the value of $a + b.$
AnswerWhen polynomial $f(x) = x^4 - 2x^3 + 3x^2 - ax - b$ divided by $(x - 1)$
The remainder is $6.$
$i. e. f(1) = 6$
$(1)^4 - 2(1)^3 + 3(1)^2 - a(1) - b = 6$
$1 - 2 + 3 - a - b = 6$
$2 - (a + b) = 6$
$(a + b) = -4$
Thus, the value of $a + b = -4.$
View full question & answer→Question 132 Marks
If $f(x) = 2x^3 - 13x^2 + 17x + 12,$ Find: $f(2)$
AnswerThe given polynomial is $f(x) = 2x^3 - 13x^2 + 17x + 12$
$f(2)$
we need to substitute the $'2\ '$ in $f(x)$
$f(2) = 2(2)^3 - 13(2)^2 + 17(2) + 12$
$= (2 \times 8) - (13 \times 4) + (17 \times 2) + 12$
$= 16 - 52 + 34 + 12$
$= 10$
therefore $f(2) = 10$
View full question & answer→Question 142 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{f(x)}=\text{l(x)}+\text{m},\text{x}=-\frac{\text{m}}{\text{l}}$
Answer$\text{f(x)}=\text{l(x)}+\text{m},\text{x}=-\frac{\text{m}}{\text{l}}$
We know that,
$\text{f(x)}=\text{l(x)}+\text{m}$
Given, that $\text{x}=-\frac{\text{m}}{\text{l}}$
Substitute the value of x in f(x)
$\text{f}\Big(-\frac{\text{m}}{\text{l}}\Big)=\text{I}\Big(-\frac{\text{m}}{\text{l}}\Big)+\text{m}$
$=-\text{m}+\text{m}$
$=0$
Since, the result is 0, $\text{x}=-\frac{\text{m}}{\text{l}}$ is the root of lx + m
View full question & answer→Question 152 Marks
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case:
$\text{f(x)}=3\text{x}+1,\text{x}=-\frac{1}{3}$
Answer$\text{f(x)}=3\text{x}+1,\text{x}=-\frac{1}{3}$
We know that,
$\text{f(x)}=3\text{x}+1$
Substitite $\text{x}=-\frac{1}{3}$ in f(x)
$\text{f}\Big(-\frac{1}{3}\Big)=3\Big(-\frac{1}{3}\Big)+1$
$= -1 + 1$
$=0$
Since, the result is 0 $\text{x}=-\frac{1}{3}$ is the root of 3x + 1
View full question & answer→Question 162 Marks
In the following two polynomials, find the value of $a,$ if $x - a$ is factor: $x^6 - ax^5 + x^4 - ax^3 + 3x - a + 2$
Answer$x^6 - ax^5 + x^4 - ax^3 + 3x - a + 2$
Let,
$x - a = 0$
$\Rightarrow x = a$
$\because x - a$ is a factor of the polynomial $f(x)$
$\therefore f(a) = 0$
$f(a) = 0$
$\Rightarrow a^6 - a \times a^5 + a^4 - a \times a^3 + 3a - a + 2 = 0$
$\Rightarrow a^6 - a^6 + a^4 - a^4 + 3a - a + 2 = 0$
$\Rightarrow 2a + 2 = 0$
$\Rightarrow 2a = -2$
$\Rightarrow a = -1$
View full question & answer→Question 172 Marks
In the following, use factor theorem to find whether polynomial $g(x)$ is a factor of polynomial $f(x)$ or, not:
$f(x) = x^3 - 6x^2 + 11x - 6; g(x) = x - 3$
AnswerLet $g(x) = 0$
$\Rightarrow x - 3 = 0$
$\Rightarrow x = 3$
$f(3) = 3^3 - 6(3)^2 + 11(3) - 6$
$= 27 - 6(9) + 33 - 6$
$= 60 - 60 = 0$
$\because f(3) = 0,$ by factor theorem $x - 3$ is a factor of $f(x).$
View full question & answer→Question 182 Marks
In the following two polynomials, find the value of $a,$ if $x + a$ is a factor. $x^4 - a^2x^2 + 3x - a$
Answer$x^4 - a^2x^2 + 3x - a$
Let,
$x + a = 0$
$\Rightarrow x = -a$
$\because (x + a)$ is a factor of $f(x) = x^4 - a^2x^2 + 3x - a$
$\therefore f(-a) = 0$
$f(-a) $
$\Rightarrow (-a)^4 - a^2(-a)^2 + 3(-a) - a = 0$
$\Rightarrow a^4 + a^4 - 3a - a = 0$
$\Rightarrow -4a = 0$
$\Rightarrow a = 0$
View full question & answer→Question 192 Marks
Find the remainder when $x^3 + 3x^3 + 3x + 1$ is divided by: $\text{x}-\frac{1}{2}$
AnswerHere$, f(x) = x^3 + 3x^2 + 3x + 1$
By remainder theorem
$\text{x}-\frac{1}{2}$
$\Rightarrow\ \text{x}=\frac{1}{2}$
substitute the value of $x$ in $f(x)$
$\text{f}\Big(\frac{1}{2}\Big)=\Big(\frac{1}{2}\Big)^3+3\Big(\frac{1}{2}\Big)^2+3\Big(\frac{1}{2}\Big)+1$
$=\Big(\frac{1}{2}\Big)^3+3\Big(\frac{1}{2}\Big)^2+3\Big(\frac{1}{2}\Big)+1$
$=\frac{1}{8}+\frac{3}{4}+\frac{3}{2}+1$
$=\frac{1+6+12+8}{8}$
$=\frac{27}{8}$
View full question & answer→Question 202 Marks
If $f(x) = 2x^3 - 13x^2 + 17x + 12,$ Find: $f(0)$
AnswerThe given polynomial is $f(x) = 2x^3 - 13x^2 + 17x + 12$
$f(0)$
we need to substitute the $'(0)'$ in $f(x)$
$f(0) = 2(0)^3 - 13(0)^2 + 17(0) + 12$
$= (2 \times 0) - ( 13 \times 0) + (17 \times 0) + 12$
$= 0 - 0 + 0 + 12$
$= 12$
therefore $f(0) = 12$
View full question & answer→Question 212 Marks
In the following two polynomials, find the value of $a,$ if $x - a $is factor: $x^5 - a^2x^3 + 2x + a + 1$
Answer$x^5 - a^2x^3 + 2x + a + 1$
Let,
$x - a = 0$
$\Rightarrow x = a$
$\because (x - a)$ is a factor of $f(x) = x^5 - a^2x^3 + 2x + a + 1$
$\therefore f(a) = 0$
$f(a)$
$\Rightarrow a^5 - a^2 \times a^3 + 2a + a + 1 = 0$
$\Rightarrow a^5 - a^5 + 3a + 1$
$\Rightarrow 3a = -1$
$\Rightarrow\ \text{a}=-\frac{1}{3}$
View full question & answer→Question 222 Marks
Give one example each of a binomial of degree $35,$ and of a monomial of degree $100$
AnswerGiven, to write the examples for binomial and monomial with the given degrees.
Example of a binomial with degree $35 - 7x^{35} - 5$
Example of a monomial with degree $100 - 2t^{100}$
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