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),
Question 3
Find the remainder (without division) on dividing f(x) by (2x + 1) where
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
∴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)
∵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,
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 $2 x^{2}+5 x – 3$.
$f(x)=2 x^{2}+5 x-3=2(-3)^{2}+5(-3)-3$
$f(-3)=18-15-3=0$
∵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 $3 x^{2}-x-10$.
∵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),
∵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),
∵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,
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
substituting the value of x in f(x),
$f\left(-\frac{7}{2}\right)=2\left(-\frac{7}{2}\right)^{3}+5\left(-\frac{7}{2}\right)^{2}-11\left(-\frac{7}{2}\right)-14$
$=\frac{-343}{4}+\frac{245}{4}+\frac{77}{2}-14$
$=\frac{-343+245+154-56}{4}=\frac{-399+399}{4}=0$
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.
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)=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
∴$f(x)=(x-1)\left(x^{2}+11 x-26\right)$
$=(x-1)\left(x^{2}+13 x-2 x-26\right)$
=(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
$f\left(-\frac{1}{2}\right)=6\left(-\frac{1}{2}\right)^{3}+5\left(-\frac{1}{2}\right)^{2}+a\left(-\frac{1}{2}\right)-2$
$=6\left(-\frac{1}{8}\right)+5\left(\frac{1}{4}\right)+a\left(-\frac{1}{2}\right)-2$
$=-\frac{3}{4}+\frac{5}{4}-\frac{a}{2}-2$
$=\frac{-3+5-2 a-8}{4}=\frac{-6-2 a}{4}$
∵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 :
$f\left(\frac{2}{3}\right)=3\left(\frac{2}{3}\right)^{3}-k\left(\frac{2}{3}\right)^{2}+21\left(\frac{2}{3}\right)-10$
$=3 \times \frac{8}{27}-k \times \frac{4}{9}+21 \times \frac{2}{3}-10$
$=\frac{8}{9}-\frac{4 k}{9}+14-10=\frac{8-4 k}{9}+4$
∵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) = $2 x^{3}-x^{2}+p x-2$
$=2 \times 8-4+2 p-2$
=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)
$\Rightarrow-2 x^{2}+4 x+6=0$
$\Rightarrow x^{2}-2 x-3=0 \quad$ (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
$\Rightarrow 2 x^{3}-5 x^{2}+5 x-k=0$
$\Rightarrow 2\left(\frac{3}{2}\right)^{3}-5\left(\frac{3}{2}\right)^{2}+5\left(\frac{3}{2}\right)-k=0$
$\Rightarrow 2 \times \frac{27}{8}-5 \times \frac{9}{4}+5 \times \frac{3}{2}-k=0$
$ \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
∴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
$f\left(\frac{-3}{2}\right)=a\left(\frac{-3}{2}\right)^{3}+3\left(\frac{-3}{2}\right)^{2}+b\left(\frac{-3}{2}\right)-3$
$=a\left(\frac{-27}{8}\right)+3\left(\frac{9}{4}\right)+b\left(\frac{-3}{2}\right)-3$
$=\frac{-27 a}{8}+\frac{27}{4}-\frac{3 b}{2}-3$
∵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
∴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)
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