Question 12 Marks
Give an example of a binomial of degree $8.$
AnswerA polynomial having two term is called a binomial. Since the degree of required binomial is $8,$ so the highest power of $x$ in the binomial should be $8.$
An example of a binomial of degree $8$ is $2x^8 - 3x.$
View full question & answer→Question 22 Marks
Using factor theorem, show that g(x) is a factor of p(x), when $\text{p}(\text{x})=2\sqrt2\text{x}^2+5\text{x}+\sqrt2,$ $\text{g}(\text{x})=\text{x}+\sqrt2$
Answer $\text{p}(\text{x})=2\sqrt2\text{x}^2+5\text{x}+\sqrt2,$By the factor theorem, (x - a) will be factor of p(x) if p(a) = 0.
Here, $\text{f}(-\sqrt{2})=2\sqrt2(-\sqrt2)^2+5(-\sqrt2)+\sqrt2$
$=2\sqrt2\times2-5\sqrt2+\sqrt2$
$=4\sqrt2-5\sqrt2+\sqrt2$
$=5\sqrt2-5\sqrt2=0.$
$\therefore(\text{x}+\sqrt2)$ is a factor of $2\sqrt2\text{x}^2+5\text{x}+\sqrt2.$
View full question & answer→Question 32 Marks
Give an example of a monomial of degree 0.
AnswerA polynomial having one term is called a monomial. Since the degree of required monomial is 0, so the highest power of x in the monomial should be 0.
An example of a monomial of degree 0 is 5.
View full question & answer→Question 42 Marks
Find the value of a for which $(x + 1)$ is a factor of $(ax^3 + x^2 - 2x + 4a - 9).$
AnswerLet $p(x) = ax^3 + x^2 - 2x + 4a - 9$
It is given that $(x + 1)$ is a factor of $p(x).$
$\Rightarrow p(-1) = 0$
$\Rightarrow a(-1)^3 + (-1)^2 - 2(-1) + 4a - 9 = 0$
$\Rightarrow -a + 1 + 2 + 4a - 9 = 0$
$\Rightarrow 3a - 6 = 0$
$\Rightarrow 3a = 6$
$\Rightarrow a = 2$
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Verify that: $1$ and $2$ are the zeros of the polynomial $p(x) = x^2 - 3x + 2.$
Answer$p(x) = x^2 - 3x + 2 = (x - 1)(x - 2)$
$\Rightarrow p(1) = (1 - 1) \times (1 - 2)$
$= 0 \times (-1)$
$= 0$
Also,
$p(2) = (2 - 1)(2 - 2)$
$= (-1) \times 0$
$= 0$
Hence, $1$ and $2$ are the zeroes of the given polynomial.
View full question & answer→Question 62 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x),$ when
$p(x) = 2x^4 + 9x^3 + 6x^2 - 11x - 6, g(x) = x - 1$
Answer$f(x) = (2x^4 + 9x^3 + 6x^2 - 11x - 6)$
By the Factor Theorem, $(x - 1)$ will be a factor of $f(x)$ if $f(1) = 0.$
Here, $f(1) = 2 \times 1^4 + 9 \times 1^3 + 6 \times 1^2 - 11 \times 1 - 6$
$= 2 + 9 + 6 - 11 - 6$
$= 17 - 17 = 0$
$\therefore (x - 1)$ is factor of $(2x^4 + 9x^3 + 6x^2 - 11x - 6).$
View full question & answer→Question 72 Marks
Give an example of a trinomial of degree $4.$
AnswerA polynomial having three term is called a trinomial. Since the degree of required binomial is $4,$ so the highest power of $x$ in the trinomial should be $4.$
An example of a trinomial of degree $4$ is $2x^4 - 3x + 5.$
View full question & answer→Question 82 Marks
Find the value of a for which $(x - 4)$ is a factor of $(2x^3 - 3x^2 - 18x + a).$
Answer$f(x) = (2x^3 - 3x^2 - 18x + a)$
$x - 4 = 0$
$\Rightarrow x = 4$
$\therefore f(4) = 2(4)^3 - 3(4)^2 - 18 \times 4 + a$
$= 128 - 48 - 72 + a$
$= 128 - 120 + a$
$= 8 + a$
Given that $(x - 4)$ is a factor of $f(x).$
By the Factor Theorem, $(x - a)$ will be a factor of f$(x)$ if $f(a) = 0$ and therefore $f(4) = 0.$
$\Rightarrow f(4) = 8 + a = 0$
$\Rightarrow a = -8$
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Using factor theorem, show that $g(x)$ is a factor of $p(x),$ when
$p(x) = 2x^3 + 7x^2 - 24x - 45, g(x) = x - 3$
Answer$f(x) = (2x^3 + 7x^2 - 24x - 45)$
By the Factor Theorem, $(x - 3)$ will be a factor of $f(x)$ if $f(3) = 0.$
Here, $f(3) = 2 \times 3^3 + 7 \times 3^2 - 24 \times 3 - 45$
$= 54 + 63 - 72 - 45$
$= 117 - 117 = 0$
$\therefore (x - 3)$ is a factor of $(2x^3 + 7x^2 - 24x - 45).$
View full question & answer→Question 102 Marks
Find the value of a for which $(x + 2a)$ is a factor of $(x^5 - 4a^2x^3 + 2x + 2a + 3).$
AnswerLet $p(x) = x^5 - 4a^2x^3 + 2x + 2a + 3$
It is given that $(x + 2a)$ is a factor of $p(x).$
$\Rightarrow p(-2a) = 0$
$\Rightarrow (-2a)^5 - 4a^2(-2a)^3 + 2(-2a) + 2a + 3 = 0$
$\Rightarrow -32a^5 - 4a^2(-8a^3) - 4a + 2a + 3 = 0$
$\Rightarrow -32a^5 + 32a^5 -2a + 3 = 0$
$\Rightarrow 2a = 3$
$\Rightarrow\text{a}=\frac{3}{2}$
View full question & answer→Question 112 Marks
Find the value of a for which the polynomial $(x^4 - x^3 - 11x^2 - x + a)$ is divisible by $(x + 3).$
AnswerLet $p(x) = x^4 - x^3 - 11x^2 - x + a$
It is given that $p(x)$ is divisible by $(x + 3).$
$\Rightarrow (x + 3)$ is a factor of $p(x).$
$\Rightarrow p(-3) = 0$
$\Rightarrow (-3)^4 - (-3)^3 - 11(-3)^2 - (-3) + a = 0$
$\Rightarrow 81 + 27 - 99 + 3 + a = 0$
$\Rightarrow 12 + a = 0$
$\Rightarrow a = -12$
View full question & answer→Question 122 Marks
Find the value of $k$ for which $(x - 1)$ is a factor of $(2x^3 + 9x^2 + x + k).$
Answer$f(x) = (2x^3 + 9x^2 + x + k)$
$x - 1 = 0$
$\Rightarrow x = 1$
$\therefore f(1) = 2 \times 1^3 + 9 \times 1^2 + 1 + k$
$= 2 + 9 + 1 + k$
$= 12 + k$
Given that $(x - 1)$ is a factor of $f(x).$
By the Factor Theorem, $(x - a)$ will be a factor of $f(x)$ if $f(a) = 0$ and therefore $f(1) = 0.$
$\Rightarrow f(1) = 12 + k = 0$
$\Rightarrow k = -12.$
View full question & answer→Question 132 Marks
Rewrite the following polynomial in standard form. $x - 2x^2 + 8 + 5x^3$
Answer$8 + x - 2x^2 + 5x^3$ is a polynomial in standard form as the powers of $x$ are in ascending order.
View full question & answer→Question 142 Marks
Verify that:
$0$ and $3$ are the zeros of the polynomial, $r(x) = x^2 - 3x.$
Answer$r(x) = x^2 - 3x$
$r(0) = 0^2 - 3 \times 0$
Also,
$r(3) = 3^2 - 3 \times 3$
$= 9 - 9$
$= 0$
Hence, $0$ and $3$ are the zeroes of the given polynomial.
View full question & answer→Question 152 Marks
Using factor theorem, show that g(x) is a factor of p(x), when
$\text{p}(\text{x})=7\text{x}^2-4\sqrt2\text{x}-6,\ $ $\text{g}(\text{x})=\text{x}-\sqrt2$
Answer$\text{p}(\text{x})=7\text{x}^2-4\sqrt2\text{x}-6$
By the factor theorem, (x - a) will be factor of p(x) if(a) = 0.
Here, $\text{f}(\sqrt{2})=7(\sqrt{2})^2-4\sqrt{2}\times\sqrt{2}-6$
$=14-8-6$
$=14-14=0$
$\therefore(\text{x}-\sqrt2)$ is a factor of $\big(7\text{x}^2-4\sqrt2\text{x}-6\big).$
View full question & answer→Question 162 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x),$ when
$p(x) = 2x^3 + 9x^2 - 11x - 30, g(x) = x + 5$
Answer$f(x) = 2x^3 + 9x^2 - 11x - 30$
By the Factor Theorem, $(x + 5)$ will be a factor of $f(x)$ if $f(-5) = 0.$
Here, $f(-5) = 2(-5)^3 + 9(-5)^2 - 11(-5) - 30$
$= -250 + 225 + 55 - 30$
$= -280 + 280 = 0$
$\therefore (x + 5)$ is a factor of $(2x^3 + 9x^2 - 11x - 30).$
View full question & answer→Question 172 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x),$ when
$p(x) = 69 + 11x - x^2 + x^3, g(x) = x + 3$
AnswerBy the factor theorem, $g(x) = x + 3$ will be a factor of $p(x)$ if $p(-3) = 0.$
Now, $p(x) = 69 + 11x - x^2 + x^3$
$\Rightarrow p(-3) = 69 + 11(-3) - (-3)^2 + (-3)^3$
$= 69 - 33 - 9 - 27$
$= 0$
Hence, $g(x) = x + 3$ is a factor of the given polynomial $p(x).$
View full question & answer→Question 182 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x),$ when
$p(x) = x^4 - x^2 - 12, g(x) = x + 2$
Answer$f(x) = (x^4 - x^2 - 12)$
By the Factor Theorem, $(x + 2)$ will be a factor of $f(x)$ if $f(-2) = 0.$
Here, $4$
$= 16 - 4 - 12$
$= 16 - 16 = 0$
$\therefore (x + 2)$ is a factor of $4$
View full question & answer→Question 192 Marks
Verify that: $2$ and $-3$ are the zeros of the polynomial $q(x) = x^2 + x - 6.$
Answer$q(x) = x^2 + x - 6$
$\Rightarrow q(2) = 2^2 + 2 - 6$
$= 4 - 4$
$= 0$
Also,
$q(-3) = (-3)^2 + (-3) - 6$
$= 9 - 9$
$= 0$
Hence, $2$ and $-3$ are the zeroes of the given polynomial.
View full question & answer→Question 202 Marks
Using factor theorem, show that $g(x)$ is a factor of $p(x),$ when
$p(x) = x^3 - 8, g(x) = x - 2$
Answer$f(x) = (x^3 - 8)$
By the Factor Theorem, $(x - 2)$ will be a factor of $f(x)$ if $f(2) = 0.$
Here, $f(2) = (2)^3 - 8$
$= 8 - 8 = 0$
$\therefore (x - 2)$ is a factor of $(x^3 - 8).$
View full question & answer→Question 212 Marks
Give an example of a monomial of degree $5.$
AnswerA polynomial having one term is called a monomial. Since the degree of required monomial is $5,$ so the highest power of $x$ in the monomial should be $5.$
An example of a monomial of degree $5$ is $2x^5$.
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