[2 Mark Question Answer]

Question 12 Marks

Find the value of m ·when $x^3 + 3x^2 -m x +4$ is exactly divisible by $(x-2)$

Answer

$x - 2 = 0 \Rightarrow x = 2$ and remainder is $0$
Substituting this value , we get :
$f (2) = 2 \times 2 \times 2 + 3 \times 2 \times 2 - m \times 2 + 4 = 0$
$\Rightarrow 2m = 24$
$\Rightarrow m = 12$

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Question 22 Marks

Using remainder theorem, find the value of m if the polynomial $f(x)= x^3 + 5x^2 -mx +6$ leaves a remainder $2\ m$ when divided by $(x-1)$,

Answer

$x - 1 = 0 \Rightarrow x = 1$ and remainder is $2 m$
Substituting this value, we get :
$f(x) = 1 \times 1 \times 1 + 5 \times 1 \times 1 - m \times 1 + 6 = 2m$
$\Rightarrow 3m = 12$
$\Rightarrow m = 4$

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Question 32 Marks

Prove by factor theorem that
$(x - 3)$ is a factor of $5x^2 - 21 x +18$

Answer

$x - 3 = 0 \Rightarrow x = 3$
Substituting this value , we get
$f(3) = 5(3)^2 - 21(3) + 18$
$= 45 - 63 + 18$
$= 0$

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Question 42 Marks

Prove by factor theorem that
$(x-2)$ is a factor of $2x^3- 7x -2$

Answer

$(x-2)$ is a factor of $2x^3- 7x -2$
$x - 2 = 0 \Rightarrow x = 2$
Substituting this value, we get
$f\{2)=2 \times 2 \times 2 \times 2-7 \times 2-2=0$
Hence $(x-2)$ is a factor of $2 \times 3-7 x-2$

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Question 52 Marks

Prove that $(x - y)$ is a factor of $yz( y^2 - z^2) + zx( z^2 - x^2) + xy ( x^2 - y^2)$

Answer

If $x - y$ is assumed to be fsctor, then $x = y$. Substituting this in problerr polynomial, we get :
$f(x = y) = yz (y^2 - z^2) + zy(z^2 - y^2) + yy (y^2 - y^2)$
$= yz (y^2-z^2) + zy(-(y^2 - z^2)) + 0$
$= yz (y^2 - z^2) - yz (y^2 - z^2) = 0$
Hence , $(x - y)$ is a factor.

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Question 62 Marks

Prove that $( p-q)$ is a factor of $(q - r)^3 + (r - p) ^3$

Answer

If $p - q$ is assumed to be factor, then $p = q$. Substituting this in problem polynomial, we get:
$f(p = q) = (p - r)^3 + (r - p )^3$
$= (p-r)^3+ (- (p - r))^3$
$= (p - r)^3 - (p - r)^3$
$= 0$
Hence, $(p - q)$ is a factor.

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Question 72 Marks

Find without division, the remainder in the following:
$2x^3 - 3x^2 + 6x - 4$ is divisible by $(2x-3)$

Answer

$2 x^3-3 x^2+6 x-4$ is divisible by $(2 x-3)$
Putting $2 x-3=0$, we get $: x=\frac{3}{2}$
Substituting this value of $x$ in the equation, we get
$2 \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}-3 \times \frac{3}{2} \times \frac{3}{2}+6 \times \frac{3}{2}-4 $
$=\frac{27}{4}-\frac{27}{4}+9-4$
$=5$

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Question 82 Marks

Find without division, the remainder in the following :
$x^3 + 8x^2 + 7x- 11$ is divisible by $(x+4)$

Answer

$x^3 + 8x^2 + 7x- 11$ is divisible by $(x+4)$
Putting $x + 4 = 0$, we get : $x = -4$
Substituting this value of $x$ in the equation, we get
$( -4) \times (-4) \times (-4) + 8 \times ( -4 ) \times ( -4) + 7\times ( -4) - 11$
$= - 64 + 128 - 28 - 11$
$= 25$

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Question 92 Marks

Find without division, the remainder in the following:
$8x^2 - 2x + 1$ is divided by $(2x+ 1)$

Answer

$8 x^2-2 x+1$ is divided by $(2 x+1)$
Putting $2 x+1=0$, we get : $x=-\frac{1}{2}$
Substituting this value of $x$ in the equation, we get
$8 \times\left(-\frac{1}{2}\right) \times\left(-\frac{1}{2}\right)-2 \times\left(-\frac{1}{2}\right)+1$
$=2+1+1$
$=4$

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Question 102 Marks

What number should be added to $2x^3 - 3x^2 + 7x -8$ so that the resulting polynomial is exactly divisible by $(x-1)$ ?

Answer

$(x - 1) = 0 \Rightarrow x = 1$
When we substitute this value in the polynomial, whatever we get as a remainder (say a) should be added so that polynomial is exactly subtracted by the factor.
$f(1) = 2 \times 1 \times 1 \times 1 - 3 \times 1 \times 1 + 7 \times 1 - 8 + a = 0$
$\Rightarrow a = 2$
Hence answer $= 2$

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Question 112 Marks

Find without division, the remainder in the following:
$5x^3 - 7x^2 +3$ is divided by $(x-1)$

Answer

$5x^3 - 7x^2 +3$ is divided by $(x-1)$
Putting $x -1=0$, we get: $x = 1$
Substituting this value of $x$ in the equation, we get
$5 \times 1 \times 1 \times 1 - 7 \times 1 \times 1 + 3$
$= 5 - 7 + 3$
$= 1$

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Question 122 Marks

What number should be subtracted from $x^2 + x + 1$ so that the resulting polynomial is exactly divisible by $(x-2)$ ?

Answer

$(x - 2) = 0 \Rightarrow x = 2$
When we substitute this value in the polynomial, whatever we get as a remainder (say a) should be subtracted so that polynomial is exactly subtracted by the factor.
$f(2) = 2 \times 2 + 2 + 1 - a = 0$
$\Rightarrow a = 7$
Hence answer $= 7$

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Question 132 Marks

Find without division, the remainder in the following:
$5x^2 - 9x + 4$ is divided by $(x - 2)$

Answer

$5x^2 - 9x + 4$ is divided by $(x - 2)$
Putting $x - 2 = 0$, we get : $x = 2$
Substituting this value of x in the equation,
we get $5 \times 2 \times 2 - 9 \times 2 + 4 = 20 - 18 + 4 = 6$

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