[2 Mark Question Answer]

Question 12 Marks

If the polynomials $a x^3+4 x^2+3 x-4$ and $x^3-4 x+$ a leave the same remainder when divided by $(x-3)$, find the value of $a$.

Answer

Let $p(x) = ax^3 + 4x^2+ 3x - 4$ and $q(x) = x^3 - 4x + a$ be the given polynomials.
When p(x) and q(x) are divided by (x - 3) the remainder are p(3) and q(3) respectively.
p(3) = q(3) given
$a(3)^3 + 4(3)^2 + 3 x 3 - 4 = 3^3- 4 x 3 + a$
$\Rightarrow 27a + 36 + 9 - 4 = 27 - 12 + a$
$\Rightarrow 26a = 15 - 41$
$\Rightarrow 26a = -26$
$\therefore a=-\frac{26}{26}$
$= -1.$

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Question 22 Marks

Find the remainder when the polynomial $f(x) = 2x^4- 6x^3 + 2x^2 - x + 2$ is divided by $x + 2.$

Answer

If $x + 2 = 0$
$x = -2$
$f(x) = 2x^4 - 6x^3 + 2x^2 - x + 2, ..$.[By remainder theorem]
$f(-2) = 2(-2)^4 - 6(-2)^3 + 2(-2)^2 - (-2) + 2$
$= 2(16) -6(-8) + 2(4) + 2 + 2$
$= 32 + 48 + 8 + 2 + 2 = 92$
Hence, required remainder $= 92.$

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Question 32 Marks

Use the factor theorem to factorise completely $x^3 + x^2 - 4x - 4$.

Answer

$\mathrm{x} 3+\mathrm{x} 2-4 \mathrm{x}-4$
$\text { Let } \mathrm{x}+1=0$
$\therefore \mathrm{x}=-1$
On substituting value of $x$ in the expression
$\therefore f(-1)=(-1)^3+(-1)^2-4(-1)-4=0$
Clearly $x+1$ is a factor of
$f(x)=x^3+x^2-4 x-4$
$\therefore f(x)=(x+1)\left(x^2-4\right) \ldots($ By actual division)
$=(x+1)(x-2)(x+2)$

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Question 42 Marks

Use the factor theorem to determine that $x - 1$ is a factor of $x^6 - x^5 + x^4 - x^3 + x^2 - x + 1.$

Answer

Let $f(x) = x^6 - x^5 + x^4 - x^3+ x^2- x + 1$ to check whether $x - 1$ is a factor of$ x^6 - x^5 + x^4- x^3 + x^2 - x + 1$ we find $f(1).$
Put$ x = 1$ in equation $(i)$ we get
$f(1) = (1)^6- (1)^5 + (1)^4- (1)^3 + (1)^2 - (1) + 1$
$= 1 - 1 + 1 - 1 + 1 - 1 + 1$
$= 4 - 3$
$= 1.$
Since, $f(1) \neq 0$, So by factor theorem $(x - 1)$ is not a factor of $f(x).$

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Question 52 Marks

In the following problems use the factor theorem to find if g(x) is a factor of $p(x):$
$p(x) = x^3 + x^2 + 3x + 175$ and $g(x) = x + 5.$

Answer

$p(x) = x^3 + x^2 + 3x + 175$ and $g(x) = x + 5 ...(i)$
To check whether $(x + 5)$ is a factor of $p(x),$ we have to find p(-5), put $x = -5$ in equation $(i),$ we get
$p(-5) = (-5)^3+ (-5)^2 + 3(-5) + 175$
$= -125 + 25 - 15 + 175$
$= -140 + 200 = 60$
Since, $p(-5) \neq 0$, so by factor theorem $(x + 5)$ is not a factor of$ p(x).$

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Question 62 Marks

In the following problems use the factor theorem to find if $g(x)$ is a factor of $p(x)$ : $p(x)=2 x^3+4 x+6$ and $g(x)=x+1$

Answer

$p(x) = 2x^3 + 4x + 6$ and $g(x) = x + 1$
Now put $x = -1$ in equation (i), we get
$p(-1) = 2(-1)^3 + 4(-1) + 6$
$= 2 x - 1 - 4 + 6$
$= -2 - 4 + 6$
$= -6 + 6 = 0$
Since, $p(-1) = 0,$ so by factor theorem $(x = 1)$ is a factor of $p(x).$

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Question 72 Marks

In the following problems use the factor theorem to find if $g(x)$ is a factor of $p(x):$
$p(x) = x^3 - 3x^2 + 4x - 4$ and $g(x) = x - 2$

Answer

$p(x) = x^3- 3x^2 + 4x - 4$ and $g(x) = x - 2$
To check whether $x - 2$ is a factor of p(x) now put $x = 2$ in equation (i), we get
$p(2) = (2)^3 - 3(2)^2 + 4(2) -4$
$= 8 - 3 x 4 + 8 - 4$
$= 8 - 12 + 8 - 4$
$= 16 - 16 = 0$
Since, $p(2) = 0$, so by factor theorem $(x - 2)$ is a factor of $p(x).$

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Question 82 Marks

If $p(x) = 4x^3 - 3x^2 + 2x - 4$ find the remainderwhen p(x) is divided by:
$x + 2$

Answer

$p(x) = 4x^3 - 3x^2 + 2x - 4 ...(i)$
By the remainder theorem the required remainder $= p(2).$
Put $x = -2$ in equation $(i)$, we get
$p(-2) = 4(-2)^3 -3(-2)^2 + 2(-2)-4$
$= 4 x (-8) -3 x 4 -4 -4$
$= -32 -12 -4 -4$
$= -52$
Hence, the remainder is $-52.$

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Question 92 Marks

If $p(x) = 4x^3 - 3x^2 + 2x - 4$ find the remainderwhen $p(x)$ is divided by:
$x - 4$

Answer

$p(x) = 4x^3- 3x^2 + 2x - 4 ...(i)$
Put $x = 4$ in equation $(i)$, we get
$p(4) = 4(4)^3 - 3(4)^2 + 2(4) - 4$
$= 4 x 64 -3 x 16 + 8 - 4$
$= 256 - 48 + 8 - 4$
$= 264 - 52$
$= 212$
Hence, the remainder is $212.$

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Question 102 Marks

In the following two polynomials, find the value of ‘a’ if $x – a$ is a factor of each of the two:
$x^5 - a^2x^3+ 2x + a + 1.$

Answer

Let $p(x) = x^5 - a^2x^3+ 2x + a + 1$
Since (x -a) in a factor of p(x), so p(a) = 0.
Put x = a in equation (i) we get
$p(a) = (a)^5 - a^2 (a)^3 + 2a + a + 1 = 0$
$= a^5 - a^2 x a^3 + 3a + 1 = 0$
$= a^5 - a^5 + 3a + 1 = 0$
$= 3a + 1 = 0$
$\Rightarrow 3a = -1$
$\Rightarrow a =-\frac{1}{3}$.

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Question 112 Marks

In the following two polynomials, find the value of ‘a’ if $x – a$ is a factor of each of the two:
$x^6 - ax^5 + x^4 - ax^3 + 3a + 2$

Answer

Let $p(x) = x^6 - ax^5 + x^4 - ax^3 + 3a + 2 ...(i)$
Put $x = a$ in equation $(i)$ we get
$p(a) = (a)^6 - a(a)^5 + (a)^4 - a(a)^3+ 3(a) + 2$
$= a^6- a^6 + a^4 - a^4 + 3a + 2 = 0$
$\therefore 3a = -2$
$\therefore \quad a =\frac{-2}{3}$.

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Question 122 Marks

In the following two polynomials. Find the value of ‘a’ if x + a is a factor of each of the two:
$x^4 - a^2x^2 + 3x - a.$

Answer

Let $p(x) = x4 - a^2x^2 + 3x - a$
Put $x = -a$ in equation (i) we get
$P(-a) = (-a)^4 - a^2 (-a)^2 + 3(-a) -a$
$= a^4 - a^2 x a^2 - 3a - a = -4a$
But $p(-a) = 0$
$\Rightarrow -4a = 0$
$\Rightarrow a=\frac{0}{-4}$
$\Rightarrow a = 0.$

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