ML Aggarwal Solution Class 10 Chapter 6 Factorization Exercise 6

 Exercise 6

Question 1

Find the remainder (without divisions) on dividing f(x) by x – 2, where

(i) $f(x)=5 x^{2}-1 x+4$

(ii) $f(x)=2 x^{3}-7 x^{2}+3$

Sol :

Let x – 2 = 0, then x = 2

(i) Substituting value of x in f(x)

$f(x)=5 x^{2}-7 x+4$

$\Rightarrow f(2)=5(2)^{2}-7(2)+4$

$\Rightarrow f(2)=20-14+4=10$

Hence Remainder=10

(ii) $f(x)=2 x^{3}-7 x^{2}+3$

$\therefore f(2)=2(2)^{3}-7(2)^{2}+3=16-28+3$

Hence, Remainder=-9

Question 2

Using remainder theorem, find the remainder on dividing f(x) by (x + 3) where

(i) $f(x)=2 x^{2}-5 x+1$

(ii) $f(x)=3 x^{3}+7 x^{2}-5 x+1$

Sol :

Let x + 3 = 0

⇒ x = -3

Substituting the value of x in f(x),

(i) $f(x)=2 x^{2}-5 x+1$

$\therefore f(-3)=2(-3)^{2}-5(-3)+1$

=18+15+1=34

Hence Remainder =34 Ans.

(ii) $f(x)=3 x^{3}+7 x^{2}-5 x+1$

$=3(-3)^{3}+7(-3)^{2}-5(-3)+1$

=-81+63+15+1=-2

Hence Remainder =-2

Question 3

Find the remainder (without division) on dividing f(x) by (2x + 1) where

(i) $f(x)=4 x^{2}+5 x+3$

(ii) $f(x)=3 x^{3}-7 x^{2}+4 x+11$

Sol :

Let 2x+1=0, then $x=-\frac{1}{2}$

Substituting the value of x in f(x) :

(i) $f(x)=4 x^{2}+5 x+3$

$=4\left(-\frac{1}{2}\right)^{2}+5 \times\left(-\frac{1}{2}\right)+3$

$=4 \times \frac{1}{4}-\frac{5}{2}+3=1-\frac{5}{2}+3-4-\frac{5}{2}=\frac{3}{2}$

∴Remainder$=\frac{3}{2}$

(ii) $f(x)=3 x^{3}-7 x^{2}+4 x+11$

$=-3\left(-\frac{1}{2}\right)^{3}-7\left(-\frac{1}{2}\right)^{2}+4\left(-\frac{1}{2}\right)+11$

$=3\left(-\frac{1}{8}\right)-7\left(\frac{1}{4}\right)+4\left(-\frac{1}{2}\right)+11$

$=-\frac{3}{8}-\frac{7}{4}-2+11$

$=\frac{-3-14-16+88}{8}=\frac{55}{8}=6 \frac{7}{8}$

∴Remainder $=6 \frac{7}{8}$

Question 4

(i) Find the remainder (without division) when  $2 x^{3}-3 x^{2}+7 x-8$ is divided by x-1 (2000)

(ii) Find the remainder (without division) on dividing $3 x^{2}+5 x-9$ by (3x + 2)

Sol :

(i) Let x – 1 = 0, then x = 1

Substituting value of x in f(x)

$f(x)=2 x^{3}-3 x^{2}+7 x-8$

$=2(1)^{3}-3(1)^{2}+7(1)-8$

$=2 \times 1-3 \times 1+7 \times 1-8=2-3+7-8$

$=-2$

∴Remainder=2

(ii) Let 3x+2=0, then 3x=-2$\Rightarrow x=\frac{-2}{3}$

Substituting the value of x in f(x)

$f(x)=3 x^{2}+5 x-9$

$=3\left(-\frac{2}{3}\right)^{2}+5\left(-\frac{2}{3}\right)-9$

$=3 \times \frac{4}{9}-5 \times \frac{2}{3}-9=\frac{4}{3}-\frac{10}{3}-9$

$=-\frac{6}{3}-9=-2-9=-11$

∴Remainder=-11

Question 5

Using remainder theorem, find the value of k if on dividing $2 x^{3}+3 x^{2}-k x+ 5$ by x – 2, leaves a remainder 7. (2016)

Sol :

$f(x)=2 x^{2}+3 x^{2}-k x+5$

g(x) = x – 2, if x – 2 = 0, then x = 2

Dividing f(x) by g(x) the remainder will be

$f(2)=2(2)^{3}+3(2)^{2}-k \times 2+5$

=16+12-2k+5=33-2k

Remainder =7

∴33-2k=7

⇒33-7=2k

⇒2k=26

⇒$k=\frac{26}{2}=13$

∴k=13

Question 6

Using remainder theorem, find the value of a if the division of $x^{3}+5 x^{2}-ax + 6$ by (x – 1) leaves the remainder 2a.

Sol :

Let x – 1 = 0, then x = 1

Substituting the value of x in f(x)

$f(x)=x^{3}+5 x^{2}-a x+6$

$=(1)^{3}+5(1)^{2}-a(1)+6$

=1+5-a+6=12-a

∵Remainder =2a

∴ 12-a=2a

⇒ 12=a+2 a 

⇒ 3a=12

∴ a=4

Question 7

(i) What number must be subtracted from $2 x^{2}-5 x$ so that the resulting polynomial leaves the remainder 2, when divided by 2x + 1 ?

(ii) What number must be added to $2 x^{3}-7 x^{2}+2 x$ so that the resulting polynomial leaves the remainder – 2 when divided by 2x – 3?

Sol :

(i) Let a be subtracted from $2x^{2}-5 x$,

Dividing $2 x^{2}-5 x$ by 2x+1,

$\begin{array}{l}2x+1 \overline{)2x^2-5x-a(}x-3\\\phantom{2x+1)}2x^2+x\\\phantom{2x+1)}-\phantom{2x^2}-\phantom{x} \\\phantom{2x+1)}\overline{\phantom{2x^2}-6x-a}\\\phantom{2x+1)}\phantom{2x^2}-6x-3\\\phantom{2x+1)}\phantom{2x^2}+\phantom{6x}+\phantom{a}\\\phantom{2x+1)}\phantom{2x^2}\overline{\phantom{x}-a+3}\end{array}$

Here remainder is (3-a) but we are given that remainder is 2

∴ 3-a=2

⇒-a=2-3-1

⇒a=1

Hence 1 is to be subtracted

(ii) Let a be added to $2 x^{3}-7 x^{2}+2 x$ dividing it by 2x-3, then

$\begin{array}{l}2x-3 \overline{)2x^3-7x^2+2x+a(}x^2-2x-2\\\phantom{2x-3)}2x^3-3x^2\\\phantom{2x+1)}-\phantom{2x^3}+\phantom{3x^2} \\\phantom{2x+1)}\overline{\phantom{2x^2}-4x^2+2x}\\\phantom{2x+1)}\phantom{2x^2}-4x^2+2x\\\phantom{2x+1)}\phantom{2x^2}\underline{+\phantom{4x^2}-\phantom{2x+a-}}\\\phantom{2x+1)}\phantom{2x^2-3x^2}-4x+a\\\phantom{2x+1)}\phantom{2x^2-4x^2}-4x+a\\\phantom{2x+1)}\phantom{2x^2-4x^2}\underline{+\phantom{4x}-\phantom{a}}\\\phantom{2x+1)}\phantom{2x^2-4x^2-4}a-6\end{array}$

But remainder is -2, then 

⇒a-6=-2

⇒a=-2+6

⇒a=4

Hence 4 is to be added

Question 8

(i) When divided by x – 3 the polynomials $x^{2}-p x^{2}+x+6$ and $2 x^{3}-x^{2}-(p+3)x-6$ leave the same remainder. Find the value of ‘p’

(ii) Find ‘a’ if the two polynomials $a x^{3}+3 x^{2}-9$ and $2 x^{3}+4 x+a$, leaves the same remainder when divided by x + 3.

Sol :

By dividing

$x^{3}-p x^{2}+x+6$

and $2 x^{3}-x^{2}-(p+3) x-6$

by x-3, the remainder is same

Let x-3=0, then x=3

Now by Remainder Theorem,

Let p(x)$=x^{3}-p x^{2}+x+6$

p(3)$=(3)^{3}-p(3)^{2}+3+6$

=27-9p-9=36-9p

and q(x)$=2 x^{3}-x^{2}-(p+3) x-6$

$q(3)=2(3)^{2}-(3)^{2}-(p+3) \times 3-6$

$=2 \times 27-9-3 p-9-6$

=54-24-3p=30-3p

∵The remainder in each case is same

∴36-9p=30-3p

⇒36-30=9p-3p

⇒6=6p

⇒$p=\frac{6}{6}=1$

⇒∴p=1

(ii) Find "a" if the two polynomials $a x^{3}+3 x^{2}-9$ and $2 x^{3}+4 x+a$, leaves the same remainder when divided by x+3

The given polynomials are $a x^{3}+3 x^{2}-9$ and $2 x^{3}+4 x+a$

Let p(x)$=a x^{3}+3 x^{2}-9$

and $q(x)=2 x^{3}+4 x+a$

Given that p(x) and q(x) leave the same remainder when divided by (x+3), Thus by Remainder Theorem, we have 

p(-3)=q(-3)

$\Rightarrow a(-3)^{3}+3(-3)^{2}-9=2(-3)^{3}+4(-3)+a$

⇒-27 a+27-9=-54-12+a

⇒-27 a+18=-66+a

⇒ -27 a-a=-66-18 

⇒ -28 a=-84

⇒ $a=\frac{84}{28}$

∴a=3

Question 9

By factor theorem, show that (x + 3) and (2x – 1) are factors of $2 x^{2}+5 x – 3$.

Sol :

Let x + 3 = 0 then x = – 3

Substituting the value of x in f(x)

$f(x)=2 x^{2}+5 x-3=2(-3)^{2}+5(-3)-3$

$f(-3)=18-15-3=0$

∵Remainder=0, then x+3 is a factor

Again Let 2x-1=0, then $x=\frac{1}{2}$

Substituting the value of x in f(x)

$f(x)=2 x^{2}+5 x-3$

$f\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^{2}+5\left(\frac{1}{2}\right)-3$

$=2 \times \frac{1}{4}+\frac{5}{2}-3$

$=\frac{1}{2}+\frac{5}{2}-3=0$

∵Remainder=0

∴2x-1 is also a factor

hence proved

Question 10

Show that (x – 2) is a factor of $3 x^{2}-x-10$ Hence factorise $3 x^{2}-x-10$.

Sol :

Let x – 2 = 0, then x = 2

Substituting the value of x in f(x),

$f(x)=3 x^{2}-x-10=3(2)^{2}-2-10$

=12-2-10=0

∵Remainder is zero

∴x-2 is a factor of f(x)

Dividing $3 x^{2}-x-10$ by x-2, we get

$\begin{array}{l}x-2\overline{)3x^2-x-10(}3x+5\\\phantom{x-2)}3x^2-6x\\\phantom{x-2)}-\phantom{3x^2}+\\\phantom{x-2)}\overline{\phantom{3x^2-}5x-10}\\\phantom{x-2)}\phantom{3x^2-}5x-10\\\phantom{x-2)}\phantom{3x^2-}\underline{-\phantom{--}+}\\\phantom{x-2)3x^2-5x}\times\end{array}$

$\therefore 3 x^{2}-x-10=(x-2)(3 x+5)$

Question 11

Show that (x – 1) is a factor of $x^{3}-5 x^{2}-x+5$ Hence factorise $x^{3}-5 x^{2} – x + 5$

Sol :

Let x – 1 = 0, then x = 1

Substituting the value of x in f(x),

$f(x)=x^{3}-5 x^{2}-x+5$

$=(1)^{3}-5(1)^{2}-1+5$

=1-5-1+5=0

∵Remainder=0

∴x-1 is a factor of $x^{3}-5 x^{2}-x+5$

Now dividing f(x) by x-1, we get

$\begin{array}{l}x-1\overline{)x^3-5x^2-x+5(}x^2-4x-5\\\phantom{x-1)}x^3-x^2\\\phantom{x-1)}-\phantom{x^3}+\\\phantom{x-1)}\overline{\phantom{x^3-}-4x^2-x}\\\phantom{x-1)}\phantom{x^3-}-4x^2+4x\\\phantom{x-1)}\phantom{x^3-}+\phantom{--}-\\\phantom{x-1)x^3-}\overline{\phantom{+4x^2}-5x+5}\\\phantom{x-1)x^3-}\phantom{+4x^2}-5x+5\\\phantom{x-1)x^3-}\phantom{+4x^2}\underline{+\phantom{5x+}-\phantom{5}}\\\phantom{x-1)x^3-4x^2-5x}\times\end{array}$

$\therefore x^{3}-5 x^{2}-x+5$

$=(x-1)\left(x^{2}-4 x-5\right)$

$=(x-1)\left[x^{2}-5 x+x-5\right]$

=(x-1)[x(x-5)+1(x-5)]

=(x-1)(x+1)(x-5)

Question 12

Show that (x – 3) is a factor of $x^{3}-7 x^{2}+15 x-9$. Hence factorise $x^{3}-7 x^{2}+15 x-9$

Sol :

Let x – 3 = 0, then x = 3,

Substituting the value of x in f(x),

$f(x)=x^{3}-7 x^{2}+15 x-9=(3)^{3}-7(3)^{2}+15(3)-9$

=27-63+45-9=72-72=0

∵Remainder=0

∴x-3 is a factor of $x^{3}-7 x^{2}+15 x-9$

Now dividing it by x-3, we get

$\begin{array}{l}x-3\overline{)x^3-7x^2+15x-9(}x^2-4x+3\\\phantom{x-3)}x^3-3x^2\\\phantom{x-3)}-\phantom{x^3}+\\\phantom{x-1)}\overline{\phantom{x^3-}-4x^2-15x}\\\phantom{x-1)}\phantom{x^3-}-4x^2+12x\\\phantom{x-1)}\phantom{x^3-}+\phantom{--}-\\\phantom{x-1)x^3-}\overline{\phantom{+4x^2}+3x-9}\\\phantom{x-1)x^3-}\phantom{+4x^2}+3x-9\\\phantom{x-1)x^3-}\phantom{+4x^2}\underline{-\phantom{5x+}+\phantom{5}}\\\phantom{x-1)x^3-4x^2-5x}\times\end{array}$

$\therefore x^{3}-7 x^{2}+15 x-9$

$=(x-3)\left(x^{2}-4 x+3\right)=(x-3)\left[x^{2}-x-3 x+3\right]$

$=(x-3)[x(x-1)-3(x-1)]$

$=(x-3)(x-1)(x-3)=(x-3)^{2}(x-1)$

Question 13

Show that (2x + 1) is a factor of $4 x^{3}+12 x^{2}+11 x+3$ .Hence factorise $4 x^{3}+12 x^{2}+11 x+3$.

Sol :

Let 2x + 1 = 0,

$\operatorname{then} x=-\frac{1}{2}$

Substituting the value of x in f(x),

$f(x)=4 x^{3}+12 x^{2}+11 x+3$

$f\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^{3}+12\left(-\frac{1}{2}\right)^{2}+11\left(-\frac{1}{2}\right)+3$

$=4\left(-\frac{1}{8}\right)+12\left(\frac{1}{4}\right)+11\left(-\frac{1}{2}\right)+3$

$=-\frac{1}{2}+3-\frac{11}{2}+3$

=6-6=0

∵Remainder=0

∴2x+1 is a factor of $4 x^{3}+12 x^{2}+11 x+3$

Now dividing f(x) by 2x+1, we get

Figure to be added

$\therefore 4 x^{3}+12 x^{2}+11 x+3$

$=(2 x+1)\left(2 x^{2}+5 x+3\right)$

$=(2 x+1)\left[2 x^{2}+2 x+3 x+3\right]$

$=(2 x+1)[2 x(x+1)+3(x+1)]$

$=(2 x+1)[(x+1)(2 x+3)]$

$=(2 x+1)(x+1)(2 x+3)$ 

Question 14

Show that 2x + 7 is a factor of $2 x^{3}+5 x^{2}-11 x-14$. Hence factorise the given expression completely, using the factor theorem. (2006)

Sol :

Let 2x + 7 = 0, then 2x = -7

$x=\frac{-7}{2}$

substituting the value of x in f(x),

$f(x)=2 x^{3}+5 x^{2}-11 x-14$

$f\left(-\frac{7}{2}\right)=2\left(-\frac{7}{2}\right)^{3}+5\left(-\frac{7}{2}\right)^{2}-11\left(-\frac{7}{2}\right)-14$

$=\frac{-343}{4}+\frac{245}{4}+\frac{77}{2}-14$

$=\frac{-343+245+154-56}{4}=\frac{-399+399}{4}=0$

Hence, (2x+7) is a factor of f(x)

Now,$2 x^{3}+5 x^{2}-11 x-14=(2 x+7)\left(x^{2}-x-2\right)$

$=(2 x+7)\left[x^{2}-2 x+x-2\right]$

$=(2 x+7)[x(x-2)+1(x-2)]$

$=(2 x+7)(x+1)(x-2)$ 

Figure to be added

Question 15

Use factor theorem to factorise the following polynominals completely.

(i) $x^{3}+2 x^{2}-5 x-6$

(ii) $x^{3}-13 x-12$

Sol :

Let $f(x)=x^{3}+2 x^{2}-5 x-6$

Factors of (∵6=± 1 ;± 2,± 3,± 6)$

Let x=-1, then

$f(-1)=(-1)^{3}+2(-1)^{2}-5(-1)-6$

=-1+2(1)+5-6=-1+2+5-6=7-7=0

∵f(-1)=0

∴x+1 is a factor of f(x)

Now, dividing f(x) by x+1, we get

$f(x)=(x+1)\left(x^{2}+x-6\right)$

$=(x+1)\left(x^{2}+3 x-2 x-6\right)$

=(x+1){x(x+3)-2(x+3)}

=(x+1)(x+3)(x-2)

Figure to be added

(ii) $f(x)=x^{3}-13 x-12$

Let x=4, then

$f(x)=(4)^{3}-13(4)-12$

=64-52-12=64-64=0

∵f(x)=0

∴x-4 is a factor of f(x)

Now , dividing f(x) by (x-4) , we get

$f(x)=(x-4)\left(x^{2}+4 x+3\right)$

$=(x-4)\left(x^{2}+3 x+x+3\right)$

$=(x-4)[x(x+3)+1(x+3)]$

=(x-4)(x+3)(x+1)

Figure to be added

Question 16

(i) Use the Remainder Theorem to factorise the following expression : $2 x^{3}+x^{2}-13 x+6$. (2010)

(ii) Using the Remainder Theorem, factorise completely the following polynomial: $3 x^{2}+2 x^{2}-19 x+6$ (2012)

Sol :

(i) Let f(x) = $2 x^{3}+x^{2}-13 x+6$

Factors of 6 are ±1, ±2, ±3, ±6

Let x = 2, then

$f(2)=2(2)^{3}+(2)^{2}-13 \times 2+6$

=16+4-26+6=26-26=0

∵f(2)=0

∴x-2 is factor of f(x)

(By remainder theorem)

Dividing f(x) by x-2,  we get

Figure to be added

∴$f(x)=(x-2)\left(2 x^{2}+5 x-3\right)$

$=(x-2)\left\{2 x^{2}+6 x-x-3\right\}$

=(x-2){2 x(x+3)-1(x+3)}

=(x-2)(x+3)(2 x-1)

(ii) $P(x)=3 x^{3}+2 x^{2}-19 x+6$

$P(1)=3+2-19+6=-8 \neq 0$

$\mathrm{P}(-1)=-3+2+19+6=-24 \neq 0$

P(2)=24+8-38+6=0

Hence, (x-2) is a factor of P(x)

∴$P(x)=3 x^{3}+2 x^{2}-19 x+6$

$=3 x^{3}-6 x^{2}+8 x^{2}-16 x-3 x+6$

$=3 x^{2}(x-2)+8 x(x-2)-3(x-2)$

$=(x-2)\left(3 x^{2}+8 x-3\right)$

$=(x-2)\left(3 x^{2}+9 x-x-3\right)$

$=(x-2)\{3 x(x+3)-1(x+3)=(x-2)(x+3)(3 x-1)$

Question 17

Using the Remainder and Factor Theorem, factorise the following polynomial: $x^{3}+10 x^{2}-37 x+26$.

Sol :

$f(x)=x^{3}+10 x^{2}-37 x+26$

$f(1)=(1)^{3}+10(1)^{2}-37(1)+26$

= 1 + 10 – 37 + 26 = 0

x = 1

Figure to be added

x-1 is factor of f(x)

∴$f(x)=(x-1)\left(x^{2}+11 x-26\right)$

$=(x-1)\left(x^{2}+13 x-2 x-26\right)$

=(x-1)[x(x+13)-2(x+13)]

=(x-1)[(x-2)(x+13)]

Question 18

If (2 x + 1) is a factor of $6 x^{3}+5 x^{2}+a x-2$ find the value of a

Sol :

Let 2 x+1=0, then $x=-\frac{1}{2}$

Substituting the value of x in 

$f(x)$ $f(x)=6 x^{3}+5 x^{2}+a x-2$

$f\left(-\frac{1}{2}\right)=6\left(-\frac{1}{2}\right)^{3}+5\left(-\frac{1}{2}\right)^{2}+a\left(-\frac{1}{2}\right)-2$

$=6\left(-\frac{1}{8}\right)+5\left(\frac{1}{4}\right)+a\left(-\frac{1}{2}\right)-2$

$=-\frac{3}{4}+\frac{5}{4}-\frac{a}{2}-2$

$=\frac{-3+5-2 a-8}{4}=\frac{-6-2 a}{4}$

∵2x+1 is a factor of f(x)

∴Remainder=0

∴$\frac{-6-2 a}{4}=0 \Rightarrow-6-2 a=0$

2a=-6

a=-3

∴a=-3

Question 19

If (3x-2) is a factor of $3 x^{3}-k x^{2}+21 x-10$, find the value of k

Sol :

Let 3x-2=0, then 3x=2

Substituting the value of x in f(x) 

$f(x)=3 x^{3}-k x^{2}+21 x-10$

$f\left(\frac{2}{3}\right)=3\left(\frac{2}{3}\right)^{3}-k\left(\frac{2}{3}\right)^{2}+21\left(\frac{2}{3}\right)-10$

$=3 \times \frac{8}{27}-k \times \frac{4}{9}+21 \times \frac{2}{3}-10$

$=\frac{8}{9}-\frac{4 k}{9}+14-10=\frac{8-4 k}{9}+4$

∵Remainder=0

∴$\frac{8-4 k}{9}+4=0$

⇒8-4k+36=0

⇒-4k+44=0

⇒4k=44

∵k=11

Question 20

If (x – 2) is a factor of $2 x^{3}-x^{2}+p x-2$, then

(i) find the value of p.

(ii) with this value of p, factorise the above expression completely

Sol :

(i) Let x – 2 = 0, then x = 2

Now f(x) = $2 x^{3}-x^{2}+p x-2$

∴$f(2)=2(2)^{3}-(2)^{2}+p \times 2-2$

$=2 \times 8-4+2 p-2$

=16-4+2 p-2=10+2 p

(ii) ∴f(2)=0, then 10+2p=0

⇒2p=-10

⇒p=-5

Now , the polynomial will be

$2 x^{3}-x^{2}-5 x-2$

$=(x-2)\left(2 x^{2}+3 x+1\right)$

$=(x-2)\left[2 x^{2}+2 x+x+1\right]$

=(x-2)[2 x(x+1)+1(x+1)]

=(x-2)(x+1)(2 x+1)

Figure to be added

Question 21

Find the value of ‘K’ for which x = 3 is a solution of the quadratic equation, $(K+2) x^{2}-K x+6=0$.

Also, find the other root of the equation.

Sol :

$(K+2) x^{2}-K x+6=0$…(1)

Substitute x = 3 in equation (1)

$(-4+2) x^{2}-(-4) x+6=0$

$\Rightarrow-2 x^{2}+4 x+6=0$

$\Rightarrow x^{2}-2 x-3=0 \quad$ (Dividing by2) 

$\Rightarrow x^{2}-3 x+x-3=0$

⇒ x(x-3)+1(x-3)=0

⇒ (x+1)(x-3)=0

So, the roots are x=-1 and x=3

Thus, the other root of the equation is x=-1

Question 22

What number should be subtracted from $2 x^{3}-5 x^{2}+5 x$ so that the resulting polynomial has 2x – 3 as a factor?

Sol :

Let the number to be subtracted be k and the resulting polynomial be f(x), then

f(x) $=2 x^{3}-5 x^{2}+5 x-k$

Since, 2x – 3 is a factor of f(x),

Now, converting 2x – 3 to factor theorem

$f\left(\frac{3}{2}\right)=0$

$\Rightarrow 2 x^{3}-5 x^{2}+5 x-k=0$

$\Rightarrow 2\left(\frac{3}{2}\right)^{3}-5\left(\frac{3}{2}\right)^{2}+5\left(\frac{3}{2}\right)-k=0$

$\Rightarrow 2 \times \frac{27}{8}-5 \times \frac{9}{4}+5 \times \frac{3}{2}-k=0$

$\Rightarrow \frac{27}{4}-\frac{45}{4}+\frac{15}{2}-k=0$

⇒27-45+30-4k=0

⇒-4 k+12=0

$\Rightarrow k=\frac{-12}{-4}$

⇒k=3

Question 23

Find the value of the constants a and b, if (x – 2) and (x + 3) are both factors of the expression $x^{3}+a x^{2}+b x-12$

Sol :

Let x – 2 = 0, then x = 0

Substituting value of x in f(x)

$f(x)=x^{3}+a x^{2}+b x-12$

$f(2)=(2)^{3}+a(2)^{2}+b(2)-12$

=8+4 a+2 b-12=4 a+2 b-4

∵x-2 is a factor

∴4a+2b-4=0

4a+2b=4

2a+b=2..(i)

Again , let x+3=0, then x=-3

Substituting the value of x in f(x)

$f(x)=x^{3}+a x^{2}+b x-12$

$=(-3)^{3}+a(-3)^{2}+b(-3)-12$

=-27+9a-3b-12

=-39+9a-3b

3a-b=13...(ii)

Adding (i) and (ii)

5a=15

a=3

Substituting the value of a in (i)

2(3)+b=2

6+b=2

⇒b=2-6=-4

Hence, a=3, b=-4

Question 24

If (x + 2) and (x – 3) are factors of $x^{3}+a x+b$, find the values of a and b. With these values of a and b, factorise the given expression.

Sol :

Let x + 2 = 0, then x = -2

Substituting the value of x in f(x),

$f(x)=x^{3}+a x+b$

$f(-2)=(-2)^{3}+a(-2)+b=-8-2 a+b$

∵ x+2 is a factor

∴ Remainder is zero.

∴ -8-2a+b=0

⇒-2a+b=8

∴ 2a-b=-8...(i)

Again let x-3=0, then x=3

Substituting the value of x in f(x)

$f(x)=x^{3}+a x+b$

$f(3)=(3)^{3}+a(3)+b=27+3 a+b$

∵x-3 is a factor

∴Remainder=0

⇒27+3a+b=0

⇒3a+b=-27...(ii)

Adding (i) and (ii)

5a=-35

$\Rightarrow a=\frac{-35}{5}=-7$

Substituting value of a in (i) 

$2(-7)-b=-8$

$\Rightarrow-14-b=-8$

$-b=-8+14$

$\Rightarrow-b=6 \therefore b=-6$

Hence a=-7, b=-6

(x+2) and (x-3) are the factors of 

$x^{3}+a x+b$

$\Rightarrow x^{3}-7 x-6$

Now dividing $x^{3}-7 x-6$ by $(x+2)$

(x-3) or $x^{2}-x-6,$ we get

Figure to be added

∴Factors are (x+2), (x-3) and (x+1)

Question 25

(x – 2) is a factor of the expression $x^{3}+a x^{2}+b x+6$. When this expression is divided by (x – 3), it leaves the remainder 3. Find the values of a and b. (2005)

Sol :

As x – 2 is a factor of

$f(x)=x^{3}+a x^{2}+b x+6$

∴f(2)=0

∴$(2)^{3}+a(2)^{2}+b(2)+6=0$

⇒8+4a+2b+6=0

⇒4a+2b=-14

⇒2a+b=-7..(i)

as on dividing f(x) by x-3

remainder=3

∴f(3)=3

∴$(3)^{3}+a(3)^{2}+b(3)+6=3$

⇒27+9a+3 b+6=3

⇒9a+3 b=-30

⇒3a+b=-10...(ii)

Solving simultaneously equation (i) and (ii)

2(-3)+b=-7

∴-6+b=-7

∴b=-1

∴a=-3, b=-1

Question 26

If (x – 2) is a factor of the expression $2 x^{3}+a x^{2}+b x-14$ and when the expression is divided by (x–3), it leaves a remainder 52, find the values of a and b.

Sol :

$f(x)=2 x^{3}+a x^{2}+b x-14$

∴ (x – 2) is factor of f(x)

f(2) = 0

$2(2)^{3}+a(2)^{2}+b(2)-14=0$

16+4a+2 b-14=0 

4a+2b=-2

2a+b=-1..(i)

Also,(x-3) it leaves remainder =52

∴f(3)=52

$2(3)^{3}+a(3)^{2}+b(3)-14=52$

⇒54+9a+3b-14=52

⇒9a+3b=52-40

⇒9a+3b=12

⇒3a+b=4..(ii)

From (i) and (ii)

$\begin{aligned}2a+b=&-1\\3a+b=&4\\-\phantom{3a}-\phantom{b}=&-\phantom{4}\\ \hline -a=& -5\end{aligned}$

∴a=5 putting  (i)

∴2(5)+b=-1

b=-1-10

b=-11

∴a=5, b=-11

$\Rightarrow \frac{-27 a}{8}+\frac{27}{4}-\frac{3 b}{2}-3=0$

$\Rightarrow-27 a+54-12 b-24=0$ 

(Multiplying by 8)

$\Rightarrow-27 a-12 b+30=0$

$\Rightarrow-27 a-12 b=-30$

$\Rightarrow 9 a+4 b=10$ [dividing by (-3)]

$9 a+4 b=10$...(i)

Again let x+2=0, then x=-2

Substituting the value of x in f(x)

$f(x)=a x^{3}+3 x^{2}+b x-3$

$f(-2)=a(-2)^{3}+3(-2)^{2}+b(-2)-3$

=-8 a+12-2 b-3

=-8 a-2 b+9

∵Remainder =-3

∵-8 a-2 b+9=-3 

⇒-8a-2b=-3-9

⇒ -8a-2 b=-12 (dividing by 2)

⇒ 4a+b=6..(ii)

Multiplying (ii) by 4

$\begin{aligned}16 a+4 b &=24 \\ 9 a+4 b &=10 \\-\phantom{9 a}-\phantom{4 b}&\phantom{=}-\\\hline 7 a &=14 \\\hline \end{aligned}$

7a=14

$\Rightarrow a=\frac{14}{7}=2$

Substituting the value of a in (i)

9(2)+4b=10

18+4b=10

∴$b=\frac{-8}{4}=-2$

Here, a=2, b=-2

∴$f(x)=a x^{3}+3 x^{2}+b x-3$

$=2 x^{3}+3 x^{2}-2 x-3$

∵2x+3 is a factor

∴Dividing f(x) by x+2

Figure to be added

∴$2 x^{3}+3 x^{2}-2 x-3$

$=(2 x+3)\left(x^{2}-1\right)=(2 x+3)\left[\left(x^{2}\right)-(1)^{2}\right]$

=(2x+3)(x+1)(x-1)

Question 27

If $a x^{3}+3 x^{2}+b x-3$ has a factor (2x + 3) and leaves remainder – 3 when divided by (x + 2), find the values of a and 6. With these values of a and 6, factorise the given expression.

Sol :

Let 2x + 3 = 0 then 2x = -3

$\Rightarrow x=\frac{-3}{2}$

Substituting the value of x in f(x),

$f(x)=a x^{3}+3 x^{2}+6 x-3$

$f\left(\frac{-3}{2}\right)=a\left(\frac{-3}{2}\right)^{3}+3\left(\frac{-3}{2}\right)^{2}+b\left(\frac{-3}{2}\right)-3$

$=a\left(\frac{-27}{8}\right)+3\left(\frac{9}{4}\right)+b\left(\frac{-3}{2}\right)-3$

$=\frac{-27 a}{8}+\frac{27}{4}-\frac{3 b}{2}-3$

∵2x+3 is a factor of f(x)

∴Remainder=0

Question 28

Given $f(x)=a x^{2}+b x+2$ and $g(x)=b x^{2}+a x+1$. If x – 2 is a factor of f(x) but leaves the remainder – 15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression. $f(x) + g(x) +4 x^{2}+7 x$

Sol :

$f(x)=a x^{2}+b x+2$

$g(x)=b x^{2}+a x+1$

x – 2 is a factor of f(x)

Let x – 2 = 0

⇒ x = 2

∴$f(2)=a(2)^{2}+b \times 2+2=4 a+2 b+2$

∴4a+2 b+2=0 (∵x-2 is its factor)

⇒2a+b+1=0...(i) (Divisible by 2)

Let x-2=0⇒x=2

∴$g(2)=b(2)^{2}+a \times 2+1$

⇒4b+2a+1+15=0

⇒4b+2a+16=0

⇒2b+a+8=0 (Divisible by 2)

⇒a+2b+8=0...(ii)

Multiplying (i) by 2 and (ii) by 1

$\begin{aligned}4a+2 b+2=&0\\a+2 b+8=&0\\-\phantom{a}-\phantom{2b}-\phantom{8}\phantom{=}&\\ \hline 3a-6=&0\end{aligned}$

⇒3a=6 

$\Rightarrow a=\frac{6}{3}$

∴a=2

Substituting the value of a in (i)

⇒2×2+b+1=0

⇒4+b+1=0

⇒b+5=0

⇒b=-5

Hence, a=2, b=-5

Now , f(x)+g(x)$=4 x^{2}+7 x$

$=2 x^{2}-5 x+2+\left(-5 x^{2}+2 x+1\right)+4 x^{2}+7 x$

$=2 x^{2}-5 x+2-5 x^{2}+2 x+1+4 x^{2}+7 x$

$=6 x^{2}-5 x^{2}-5 x+2 x+7 x+2+1$

$=x^{2}+2 x+3$

$=x^{2}+x+3 x+3$

=x(x+1)+3(x+1)

=(x+1)(x+3)