[3 marks sum]

Question 13 Marks

if $x – 2$ is a factor of $x^2 + ax + b$ and $a + b = 1,$ find the values of $a$ and $b.$

Answer

Let $f(x) = x^2 + ax + b$
Since, $(x – 2)$ is a factor of $f(x).$
$\therefore $ Remainder $= f(2) = 0$
$(2)^2 + a(2) + b = 0$
$4 + 2a + b = 0$
$2a + b = -4 …(i)$
It is given that:
$a + b = 1 …(ii)$
Subtracting $(ii)$ from $(i),$ we get,
$a = -5$
Substituting the value of a in $(ii),$ we get,
$b = 1 – (-5) = 6$

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Question 23 Marks

When $x^3 + 3x^2 – mx + 4$ is divided by $x – 2,$ the remainder is $m + 3.$ Find the value of $m.$

Answer

Let $f(x)=x^3+3 x^2-m x-4$
According to the given information,
$f(2)=m+3$
$ (2)^3+3(2)^2-m(2)+4=m+3$
$ 8+12-2 m+4=m+3 $
$ 24-3=m+2 m$
$3 m=21$
$ m=7$

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Question 33 Marks

Show that $(x – 1)$ is a factor of $x^3 – 7x^2 + 14x – 8$. Hence, completely factorise the given expression.

Answer

let $f(x)=x^3-7 x^2+14 x-8$
$f(1)=(1)^3-7(1)^2+14(1)-8=1-7+14-8=0$
Hence, (x-1) is a factor of f(x)

$\begin{array}{rl}\therefore x^3-7 x^2+14 & x-8=(x-1)\left(x^2-6 x+8\right) \\ & =(x-1)\left(x^2-2 x-4 x+8\right) \\ & =(x-1)[x(x-2)-4(x-2)] \\ & =(x-1)(x-2)(x-4)\end{array}$

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Question 43 Marks

What must be subtracted from $16x^3 – 8x^2 + 4x + 7$ so that the resulting expression has
$2x + 1$ as a factor?

Answer

Let the number to be subtracted from the given polynomial be $k$
Let $f(y)=16 x^3-8 x^2+4 x+7-k$
It is given that $(2x + 1)$ is a factor of f(y).
$\therefore f\left(-\frac{1}{2}\right)=0$
$\Rightarrow 16\left(-\frac{1}{2}\right)^3-8\left(-\frac{1}{2}\right)^2+4\left(-\frac{1}{2}\right)+7-k=0$
$\Rightarrow-16 \times \frac{1}{8}-8 \times \frac{1}{4}-4 \times \frac{1}{2}+7-k=0$
$\Rightarrow -2 -2 -2+7-k = 0$
$\Rightarrow -6 + 7 - K = 0$
$\Rightarrow 1 - k = 0$
$\Rightarrow k =1$
Thus, 1 should be subtracted from the given polynomial

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Question 53 Marks

The polynomials $ax^3 + 3x^2 – 3$ and $2x^3 – 5x + a$, when divided by $x – 4$, leave the same remainder in each case. Find the value of a.

Answer

let $f(x)\left(a x^3+3 x^2-3\right)$
when $f (x)$ is dividend by $(x-4),$ remainder $=f(4)$
$f(4)=a(4)^3+3(4)^2-3=64 a+45$
let $g(x)=2 x^3-5 x+a$
When $g (x)$ is divided by $(x-4),$ remainder $=g(4)$
$g(4)=2(4)^3-5(4)+=a+108$
it is given that $f(4)=9(4)$
$64 a+45=a+108$
$63 a=63$
$a=1$

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Question 63 Marks

Using the Remainder Theorem, factorise each of the following completely $x^3 + x^2 – 4x – 4$

Answer

$f(x)=x^3+x^2-4 x-4$
$f$ or $x=1$,
$f(x)=f(-1)=(-1)^3+(-1)^2-4(-1)-4$
$\quad=-1+1+4-4=0$
Hence, $(x+1)$ is a factor of $f(x)$

$\therefore x^3+x^2-4 x-4=(x+1)\left(x^2-4\right)$
$=(x+1)(x+2)(x-2)$

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Question 73 Marks

Using the Reminder Theorem, factorise of the following completely. $2x^3 + x^2 – 13x + 6$

Answer

Let $f(x)=2 x^3+x^2-13 x+6$
For x = 2,
$f(x)=f(2)=2(2)^3+(2)^2-13(2)+6=16+4-26+6=0$
Hence, (x − 2) is a factor of f(x).

$\therefore 2 x^3+x^2-13 x+6=(x-2)\left(2 x^2+5 x-3\right)$
$=(x-2)\left(2 x^2+6 x-x-3\right)$
$=(x-2)[2 x(x+3)-(x-3)]$
$=(x-2)(x+3)(2 x-1)$

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Question 83 Marks

$(x + 5)$ is a factor of $2x^3 + 5x^2 – 28x – 15.$ Hence, factorise the expression $2x^3 + 5x^2 – 28x – 15$ completely.

Answer

Let $f(x)=2 x^3+5 x^2-28 x-15$
$x+5=0 \Rightarrow x =-5$
$\therefore \text { Remainder }= f (-5)$
$=2(-5)^3+5(-5)^2-28(-5)-15$
$=-250+125+140-15$
$=-265+265$
$=0$
Hence, $(x + 5)$ is a factoor of $f(x)$
Now, we have,

$\therefore 2 x^3+5 x^2-28 x-15=(x+5)\left(2 x^2-5 x-3\right)$
$=(x+5)\left[2 x^2-6 x+x-3\right]$
$=(x-5)[2 x(x-3)+1(x-3)]$
$=(x+5)(2 x+1)(x-3)$

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Question 93 Marks

Using the Factor Theorem, show that:$(x – 2)$ is a factor of $x^3 – 2x^2 – 9x + 18$. Hence, factorise the expression $x^3 – 2x^2 – 9x + 18$ completely.

Answer

Let $f ( x )=x^3-2 x^2-9 x+18$
$x-2=0 \Rightarrow x=2$
$\therefore$ Remainder $=f(2)$
$=(2)^3-2(2)^2-9(2)+18$
$ =8-8-18+18$
$=0$
Hence, $(x-2)$ is a factor of $f (x)$
Now, we have:

$\therefore x^3-2 x^2-9 x+18$
$=(x-2)\left(x^2-9\right)$
$=(x-2)(x+3)(x-3)$

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Question 103 Marks

Find the value of k, if $2x + 1$ is a factor of $(3k + 2)x^3 + (k − 1)$

Answer

Let $f(x)=(3 k+2) x^3+(k-1)$
$2 x+1=0 \Longrightarrow x=\frac{-1}{2}$
Since, $2 x+1$ is a factor of $f(x)$, remainder is $0.$
$\therefore(3 k+2)\left(\frac{-1}{2}\right)^3+(k-1)=0 $
$ \Rightarrow \frac{-(3 k+2)}{8}+(k-1)=0$
$\Rightarrow \frac{-3 k-2+8 k-8}{8}=0$
$ \Rightarrow 5 k -10=0$
$ \Rightarrow k =2$

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Question 113 Marks

Find ‘a‘ if the two polynomials $ax^3 + 3x^2 – 9$ and $2x^3 + 4x + a$, leave the same remainder when divided by $x + 3.$

Answer

$x+3=0 \Rightarrow x=-3$
Since, the given polynomials leave the same remainder when dividend by $(x-3),$
value of polynomial $a x^3+3 x^2-9$ at $x=-3$ is sae as value of polynomial $2 x^3+4 x+$ aat $x=-3$
$\Rightarrow a(-3)^3+3(-3)^2-9=2(-3)^3+4(-3)++a$
$\Rightarrow-27 a+27-9=-54-12+a$
$\Rightarrow-27 a+18=-66+a$
$\Rightarrow 28 a=84$
$\Rightarrow a=\frac{84}{28}$
$\Rightarrow a=3$

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Question 123 Marks

The polynomials $2x^3 – 7x^2 + ax – 6 and x^3 – 8x^2 + (2a + 1)x – 16$ leaves the same remainder when divided by $x – 2$. Find the value of ‘a’.

Answer

Lef $f(x)=2 x^3-7 x^2+a x-6$
$x-2=0$
$\Rightarrow x=2$
When f (x) is divided by (x-2), remainder = f(2)
$\begin{aligned} \therefore f(2) & =2(2)^3-7(2)^2+a(2)-6 \\ & =16-28-2 a-6 \\ & =2 a-18\end{aligned}$
Let $g(x)=x^3-8 x^2+(2 a+1) x-16$
When $g ( x )$ is dividend by $( x -2)$, remainder $= g (2)$
$\begin{aligned} \therefore g(2) & =(2)^3-8(2)^2+(2 a+1)(2)-16 \\ & =8-32+4 a+2-16 \\ & =4 a-38\end{aligned}$
By the given condition, we have:
$f(2)=g(2)$
$2a-18=4a-38$
$4a-2a=38-18$
$2a=20$
$a=10$
Thus, the value of a is $10$

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Mathematics [3 marks sum]

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