[4 marks sum]

Question 14 Marks

Find the values of $m$ and $n$ when the polynomial $f(x)=x^3-2 x^2+m x+n$ has a factor $(x+2)$ and leaves a remainder 9 when divided by $( x +1$ ).

Answer

$(x+2) \Rightarrow x =- 2 .... (i)$
$(x+l) \Rightarrow x = -1 .... (ii)$
Putting (i) in polynomial, we get
$f(-2) = (-2) \times (-2)\times (-2) - 2 \times (-2) \times (-2) + m \times (-2) + n = 0$
$\Rightarrow -8 -8 - 2m + n= 0$
$\Rightarrow n =2 m + 16 .... (iii)$
Putting (ii) in polynomial, and remainder is 9 we get
$f(-1) = (-1) \times (-1) \times (-1) - 2 \times (-1) \times (-1) + m \times (-1) + n = 9$
$\Rightarrow - 1 - 2 - m + n = 9$
$\Rightarrow m = n - 12 .....(iv)$
Combining (iii) and (iv), we get,
$n = 2 x (n - 12) + 16 ,$
$\Rightarrow n = 8$
Hence, $m = n - 12 = 8 - 12 = -4$
$m = - 4, n = 8 $

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Question 24 Marks

Find the values of $a$ and $b$ when the polynomial $f(x)=a x^3+3 x^2+b x-3$ is exactly divisible by $(2 x+3)$ and leaves a remainder $-3$ when divided by $(x+2)$.

Answer

$(2 x+3) \Rightarrow x=-\frac{3}{2} \ldots . .(i) $
$(x+2) \Rightarrow x=-2 \ldots \text { (ii) }$
putting $(i)$ in polynomial, we get
$f \left(-\frac{3}{2}\right)= a \times\left(-\frac{3}{2}\right) \times\left(-\frac{3}{2}\right) \times\left(-\frac{3}{2}\right)+3 \times\left(-\frac{3}{2}\right) \times\left(-\frac{3}{2}\right)+ b \times\left(-\frac{3}{2}\right)-3=0$
$-27 a+54-12 b-24=0 $
$\Rightarrow 27 a=-12 b+30 \ldots \text { (iii) }$
Putting $(ii)$ in polynomial, and remainder is $-3$ we get
$f(-2)=a \times(-2) \times(-2) \times(-2)+3 \times(-2) \times(-2)+b \times(-2)-3=-3$
$b=6-4 a \ldots . . \text { (iv) }$
Combining $(iii)$ and $(iv),$ we get,
$27 a=-12 \times(6-4 a)+30 $
$\Rightarrow 27 a=-72+48 a+30 $
$\Rightarrow a=2, b=6-4 \times 2=-2 $
$a=2, b=-2$

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Question 34 Marks

Find the values of $p$ and $q$ in the polynomial $f(x)= x^3 - px^2 + 14x -q$, if it is exactly divisible by $(x-1)$ and $(x-2).$

Answer

$(x - 1) \Rightarrow x = l .... (i)$
$(x - 2) \Rightarrow x = 2 .... (ii)$
Putting (i) in polynomial , we get
$f(l) = 1\times 1 \times 1 - p \times 1 \times 1 + 14 \times 1 - q = 0$
$\Rightarrow p + q = 15$
$\Rightarrow p = 15 - q$
Putting (ii) in polynomial , we get
$f(2) = 2 \times 2 \times 2 - p \times 2 \times 2 + 14 \times 2 - q = 0$
$4p + q= 36, \Rightarrow q = 36 - 4p .... (iv)$
Combining (iii) and (iv), we get,
$p = 15 - (36 - 4p)$
$\Rightarrow p= 15 - 36 + 4p$
$\Rightarrow 3p = 21$
$q = 36 - 4 \times 7 = 8$
$\Rightarrow p = 7 , q = 8$

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Question 44 Marks

A polynomial $f(x)v when divided by $(x - 1)$ leaves a remainder $3$ and when divided by $(x - 2)$ leaves a remainder of $1.$ Show that when its divided by $(x - i)(x - 2),$ the remainder is $(-2x + 5).$

Answer

$ \text { Given } f(x)=(x-1)(x-2)+(-2 x+5)$
$=\left(x^2-3 x+2\right)+(-2 x+5)$
$f(x)=x^2-5 x+7 $
Substituting $=1$
$f(x)=1-5+7=3$
when $f(x)$ is divided by $(x-1)$, remainder $=3$
substituting $x=2$
$f(x)=4-10+7=1$
when $f(x)$ is divided by $(x-2)$, remainder $=1$
$\frac{x^2-5 x+7}{x^2-3 x+2}=1 \frac{-2 x+5}{(x-1)(x-2)}$
and
when $f(x)$ is divided by $(x-1)(x-2)$, remainder $=(-2 x+5)$.

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Question 54 Marks

Prove that $(5x - 4)$ is a factor of the polynomial $f(x) = 5x^3 - 4x^2 - 5x +4$. Hence factorize It completely.

Answer

If $5 x-4$ is assumed to be factor, then $x=\frac{4}{5}$. Substituting this in problem polynomial, we get:
$ f \left(\frac{4}{5}\right)=5 \times\left(\frac{4}{5}\right) \times\left(\frac{4}{5}\right) \times\left(\frac{4}{5}\right)-4 \times\left(\frac{4}{5}\right) \times\left(\frac{4}{5}\right)-5 \times\left(\frac{4}{5}\right)+4$
$=\frac{64}{25}-\frac{64}{25}-4+4$
$=0 $
Hence $(5 x-4)$ is a factor of the polynomial.
Multiplying $(5 x-4)$ by $x^2$, we get $5 x^3-4 x^2$, hence we are left with $-5 x+4$ (and 1st part of factor as $\left.x^2\right)$
Multiplying $(5 x-4)$ by $-1 ,$ we get $-5 x+4$, hence we are left with $0$ (and $2$ nd part of factor as $-7 x$ ).
Hence complete factor is $(5 x-4)\left(x^2-1\right)$.
Further factorizing $\left(x^2-1\right)$, we get :
$\Rightarrow(x-1)(x+1)=0$
Hence answer is $(5 x-4)(x-1)(x+1)=0$

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Question 64 Marks

Prove that $(x+1)$ is a factor of $x^3-6 x^2+5 x+12$ and hence factorize it completely.

Answer

If $x+1$ is assumed to be factor, then $x=-1$. Substituting this in problem polynomial, we get:
$f(-1)=(-1) \times(-1) \times(-1)-6 \times(-1) \times(-1)+5 \times(-1)+12=0$
Hence $(x+1)$ is a factcr of the polynomial.
Multiplying $(x+1)$ by $x^2$, we get $x^3+x^2$, hence we are left with $-7 x^2+5 x+12$ (and 1st part of factor as $x^2$ ).
Multiplying $(x+1)$ by $-7 x$, we get $-7 x^2-7 x$, hence we are left with $12 x+12$ (and 2 nd part of factor as $-7 x$ ).
Multiplying $(x+1)$ by 12 , we get $12 x+12$, hence we are left with 0 (and 3 rd part of factor as 12 ).
Hence complete factor is $(x+1) (x^2-7x+12).$
Further factorizing $(x^2 - 7x + 12),$ we get:
$x^2 - 3x - 4x + 12 =O$
$\Rightarrow (x - 4)(x - 3) = 0$
Hence answer is $(x + 1)(x - 4)(x - 3) = 0 $

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Question 74 Marks

Find the values of a and b in the polynomial $f(x) = 2x^3 + ax^2 + bx + 10,$ if it is exactly divisible by $(x+2)$ and $(2x-1).$

Answer

$(x+2) \Rightarrow x=-2 \ldots\ldots(i)$
$(2 x-1) \Rightarrow x=\frac{1}{2}\ldots(ii)$
Putting $(i)$ in polynomial, we get
$f(-2)=2 \times(-2) \times(-2) \times(-2)+a \times(-2) \times(-2)+b \times(-2)+10=0$
$\Rightarrow-16+4 a-2 b+10=0$
$\Rightarrow a =\frac{ b }{2}+\frac{3}{2}\ldots(iii)$
Putting $(ii)$ in polynomial, we get
$ f \left(\frac{1}{2}\right)=2 \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)+ a \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)+ b \times\left(\frac{1}{2}\right)+10=0$
$\Rightarrow \frac{1}{4}+\frac{ a }{4}+\frac{ b }{2}+10=0$
$\Rightarrow a =-2 b -41 \ldots . . . \text { (iv) }$
Combining $(iii)$ and $(iv),$ we get,
$\frac{b}{2}+\frac{3}{2}=a=-2 b-41$
$\Rightarrow \frac{b+3}{2}=-2 b-41$
$\Rightarrow b+3=-4 b-82$
$\Rightarrow 5 b=-85$
$\Rightarrow b=-17$
and $a=-7$
$\Rightarrow a =-7, b =-17$

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Question 84 Marks

The polynomial $f(x) = ax^4 + x^3 + bx^2 - 4x + c$ has $(x + 1), (x-2)$ and $(2x - 1)$ as its factors. Find the values of $a,b,c,$ and remaining factor.

Answer

When $x+1$ is a factor, we can substitute $x=-1$ to evaluate values$\ldots(i)$
When $x -2$ is a factor, we can substitute $x =2$ to evaluate values$\ldots(ii)$
When $2 x-1$ is a factor, we can substitute $x=\frac{1}{2}$ to evaluate values$\ldots(iii)$
Substituting $(1),$ we get
$f(-1)=a \times(-1)^4+(-1)^3+b(-1)^2-4(-1)+c=0$
$\Rightarrow a+b+c=-3$
$\Rightarrow a=-b-c-3 \ldots(\text { iv })$
Substituting $(11),$ we get
$\Rightarrow f(2)=a \times(2)^4+(2)^3+b(2)^2-4(2)+c=0$
$\Rightarrow 16 a+4 b+c=0 \ldots . .(v)$
Substituting $(iii),$ we get
$\Rightarrow f \left(\frac{1}{2}\right)= a \times\left(\frac{1}{2}\right)^4+\left(\frac{1}{2}\right)^3+ b \left(\frac{1}{2}\right)^2-4\left(\frac{1}{2}\right)+c=0$
$\Rightarrow \frac{ a }{16}+\frac{ b }{4}+ c =2-\frac{1}{8}$
$\Rightarrow a +4 b +16 c =30 \ldots \text { (vi) }$
Putting $(iv)$ in $(v)$ and $(vi),$ we get:
$16 a+4 b+c=0 ; \Rightarrow 16 x(-b-c-3)+4 b+c=0$
$\Rightarrow-12 b-15 c-48=0=4 b+5 c=16$
$\Rightarrow b=-4-\frac{5 c}{4} \ldots \ldots .(\text { vii })$
$a+4 b+16 c=30 \Rightarrow(-b-c-3)+4 b+16 c=30$
$\Rightarrow 3 b+15 c=33 \ldots \text { (viii) }$
Putting $(vii)\ 1n \ (viii),$ we get,
$\Rightarrow 3 \times\left(-4-\frac{5 c}{4}\right)+15 c=33 \text {, }$
$\Rightarrow$ Solving this, we get
$\Rightarrow c=4$
Putting this value of $c$ in $(viii),$ we get:
$b=-9$
Putting this value of $c$ in $(iv),$ we get:
$a=2$
Puting values of $a, b, c$ in polynomial, we get:
$f(x)=2 x^4+x^3-9 x^2-4 x+4$
In order to find the remaining factor, lets start with $(x-2)$ as one of the factor. Then,
Multiplying $(x-2)$ by $2 x^3$, we get $2 x^4-4 x^3$, hence we are left with $5 x^3-9 x^2-4 x+4 ($and $1^{st}$ part of factor as $\left.2 x^3\right)$.
Multiplying $( x-2 )$ by $5 x^2$, we get $5 x^3-10 x^2$, hence we are left with $x 2-4 x+4\ ($and $2^{nd}$ part of factor as $\left.5 x^2\right)$.
Multiplying $(x-2)$ by $x$, we get $x^2-2 x$, hence we are left with $-2 x+4 \ ($and $3^{rd}$ part of factor as $x ).$
Multiplying $(x-2)$ by $-2 ,$ we get $-2 x+4$, hence we are left with $0\ ($and $4^{th}$ part of factor as $-2 )$.
Hence complete factor is $(x-2)\left(2 x^3+5 x^2+x-2\right)$.
Again factoring $\left(2 x^3+5 x^2+x-2\right)$ by similar method we get factors as :
$(x+1)(2 x-1)(x+2)$
Hence, $f(x)=2 x^4+x^3-9 x^2-4 x+4=(x-2)(x+1)(2 x-1)(x+2)$.

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Question 94 Marks

use the rernainder theorem to find the factors of $( a-b )^3 + (b-c )^3 + ( c-a)^3 $

Answer

We know that $( a-b )^3= a^3 - 3a^2b + 3 ab^2 - b^3 ..........(i)$
And if we put $a - b = 0 c a = b,$ and substitute this to the polynomial, we get
$f(x) = 0 + (a - c)^3 + (c - a)^3 = (a - c)^3 - (a - c)^3 = 0$
Hence$, (a - b)$ is a factor. $\Rightarrow a = b .... (ii)$
Substiruong $(1)$ in problem polynomial, we get
$f (x) = 0 + (b^3 - 3b^2c + 3bc^2 - c^2) + (c^3 - 3c^2a + 3ca^2 - a^3)$
$= - 3 b^2c + 3 bc^2 - 3ca^2 + 3ca^2 $
$= 3( -b^2c + bc^2 - ca^2 + ca^2)$
If we put $b - c = 0 \Rightarrow b = c ,$ and subsorute this $ID$ the pdynorrual, we get:
$f (b = c) , 3 (-c^2 \times c + c \times c^2 - c \times c^2 + c \times c^2) = 0$
Hence, till new factors are $3 \times (a - b) \times (b - c) ... (iii)$
Similarly if we had put $c = a,$ we would have got similar result.
So $(c - a)$ is also a factor $..... (iv)$
From $(ii), (iu),$ and $(iv),$ we get
$3(a - b)(b - c)(c - a)$ is a complete factorization of the oiven polynomial.

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Question 104 Marks

Find the values of $a$ and $b$ when the factors of the polynomial $f(x)= ax^3 + bx^2 + x a$ are $(x+3)$ and $(2x-1)$. Factorize the polynomial completely.

Answer

$(x+3) \Rightarrow x=-3\ldots(i)$
$(2 x-1) \Rightarrow x=\frac{1}{2}\ldots(ii)$
Putting $(i)$ in polynomial, we get
$f(-3)=a \times(-3) \times(-3) \times(-3)+b \times(-3) \times(-3)+(-3)-a=0$
$\Rightarrow 27 a+9 b-3-a=0$
$\Rightarrow a=\frac{9 b}{28}-\frac{3}{28} \ldots . . .(\text { iii) }$
Putting $(ii)$ in polynomial, we get
$f \left(\frac{1}{2}\right)= a \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)+ b \times\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)- a =0$
$\Rightarrow \frac{ a }{8}+\frac{ b }{4}+\frac{1}{2}- a =0$
$\Rightarrow b =\frac{7 a }{2}-2 \ldots . . . .(\text { iv) }$
Combining $(iii)$ and $(iv),$ we get,
$a=\frac{9}{28} \times\left(\frac{7 a}{2}-2\right)-\frac{3}{28}$
$\Rightarrow 56 a=63 a-42$
$\Rightarrow a=6$
$\Rightarrow b=\frac{7 \times 6}{2}-2=21-2=19$
$a=6, b=19$
Putting these values in polynomial, we get
$f(x)=6 x^3+19 x^2+x-6$
Hence, equation becomes $(x+3)(2 x-1)(3 x+2)=0$

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Mathematics [4 marks sum]

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