[5 marks sum]

Question 15 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 \quad \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 25 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 1st 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"d 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'd 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 35 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 $3 x (a - b) x (b - c) ... (iii)$
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 45 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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