ICSE Class 10 Mathematics Chapter 8 Selina Concise Solutions

1. Find the remainder when $x^2-8x+4$ is divided by $2x+1$.

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

$2x+1=0\Rightarrow x=-\frac{1}{2}$

Required remainder $=$ Value of given polynomial $x^2-8x+4$ at $x=-\frac{1}{2}$

$\text{Remainder }={\left(-\frac{1}{2}\right)}^2-8\left(-\frac{1}{2}\right)+4$

$=\frac{1}{4}+4+4$

$=8\frac{1}{4}$

2. Find the value of ' $a$ ' if the division of $a{x}^3+9x^2+4x-10$ by $x+3$ leaves a remainder of $\mathbf{5}$.

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

$x+3=0\Rightarrow x=-3$

Given, the remainder is $5$.

Thus, the value of $a{x}^3+9x^2+4x-10$ at $x=-3$ is $5$

$a(-3{)}^3+9(-3{)}^2+4(-3)-10=5$

$\Rightarrow -27a+81-12-10=5$

Then, $a=2$

3. When the polynomial $2x^3-k{x}^2+(5k-3)x-8$ is divided by $x-2$, the remainder is $14$ . Find the value of ' $k$ '.

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

$x-2=0\Rightarrow x=2$

Given, the remainder is $14$.

Therefore,

$\Rightarrow 2(2{)}^3-k(2{)}^2+(5k-3)\times 2-8=14$

$\Rightarrow 16-4k+10k-6-8=14$

$\Rightarrow 6k=12\text{ and }k=2$

4. The polynomials $3x^3-a{x}^2+5x-13$ and $(a+1)x^2-7x+5$ leave the same remainder when divided by $x-3$. Find the value of ' $a$ '.

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

$x-3=0\Rightarrow x=3$

Since, the given polynomials leave the same remainder when divided by $x-3$.

Value of polynomial $3x^3-a{x}^2+5x-13$ at $x=3$ is the same as the value of polynomial $(a+1)x^2-7x+5$ at $x=3$

$\Rightarrow 3(3{)}^3-a(3{)}^2+5\times 3-13=(a+1)(3{)}^2-7\times 3+5$

$\Rightarrow 81-9a+15-13=9a+9-21+5$

$\Rightarrow 18a=90\text{ and }a=5$

5. When $f(x)=x^3+a{x}^2-bx-8$ is divided by $x-2$, the remainder is zero and when divided by $x+1$, the remainder is $-30$. Find the values of ' $a$ ' and ' $b$ '.

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Since, $x-2=0\Rightarrow x=2$

And, given that on dividing $f(x)=x^3+a{x}^2-bx-8$ by $x-2$, the remainder $=0$

Then, $f(2)=0$

$\Rightarrow$ $2^3+a(2{)}^2-b(2)-8=0$

$\Rightarrow 8+4a-2b-8=0$

$\Rightarrow 4a-2b=0$

$\Rightarrow 2a-b=0\to (1)$

Again, given that on dividing $f(x)=x^3+a{x}^2-bx-8$ by $x+1$, the remainder $=-30$

$\Rightarrow f(-1)=-30$

$\Rightarrow (-1{)}^3+a(-1{)}^2-b(-1)-8=-30$

$\Rightarrow -1+a+b-8=-30$

$\Rightarrow a+b=-21\to (2)$

When adding $(1)$ and $(2)$, we get  $a=-7$.

When substituting the value of $a$ in $(2)$, we get $b=-14$.

6. What number should be added to $2x^3-3x^2+x$ so that when the resulting polynomial is divided by $x-2$, the remainder is $3$?

Ans:

Let the number added be $k$ so the resulting polynomial is,

$2x^3-3x^2+x+k$

Here, $x-2=0\Rightarrow x=2$

Given, when this polynomial is divided by $x-2$, the remainder $=3$

$\Rightarrow$$2(2{)}^3-3(2{)}^2+2+k=3$

Substitute the value of $x$and simplify,

$\Rightarrow 16-12+2+k=3$

$\Rightarrow k=-3$

$\therefore$ The required number to be added is $-3$.

7. Determine whether $x-1$ is a factor of $x^6-x^5+x^4+x^3-x^2-x+1$ or not ?

Ans:

Since, $x-1=0\Rightarrow x=1$

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

When given polynomial is divided by $x-1$, the remainder is,

$=(1{)}^6-(1{)}^5+(1{)}^4+(1{)}^3-(1{)}^2-(1)+1$

$=1-1+1+1-1-1+1$

$=4-3=1$, which is not equal to zero.

Therefore, $x-1$ is not a factor of the given polynomial.

8. If $x-2$ is a factor of $x^2-7x+2a$, find the value of $a$.

Ans:

Here, $x-2=0\Rightarrow x=2$

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Since, $x-2$ is a factor of polynomial $x^2-7x+2a$

$\Rightarrow \text{ Remainder }=0\Rightarrow (2{)}^2-7(2)+2a=0$$

$\Rightarrow a=5$

Therefore, the value of $a$is $5.$

9. Find the value of ‘$k$’ if $(x-2)$ is a factor of $x^3+2x^2-kx+10$. Hence, determine whether $(x+5)$ is also a factor.

Ans:

Here, $x-2$ is a factor and $x-2=0\Rightarrow x=2$

Thus, the value of given expression $x^3+2x^2-kx+10$ is zero at $x=2$.

That is, remainder $=0$

$\Rightarrow (2{)}^3+2(2{)}^2-k\times 2+10=0$

$\Rightarrow 8+8-2k+10=0$

$\Rightarrow k=13$

On substituting $k=13$, the given expression becomes $x^3+2x^2-13x+10$

Now to check whether $(x+5)$ is also a factor or not,

Find the value of the given expression for $x=-5.[\because x+5=0\Rightarrow x=-5]$

$x^3+2x^2-13x+10(\text{ at }x=-5)$

$=(-5{)}^3+2(-5{)}^2-13(-5)+10$

$=-125+50+65+10$

$=-125+125=0$

Since, the remainder is $0,(x+5)$ is a factor.

10. Given that $x+2$ and $x-3$ are factors of $x^3+ax+b$; calculate the values of $a$ and $b$.

Ans:

Given, $x+2$ is a factor of $x^3+ax+b$

$\Rightarrow (-2{)}^3+a(-2)+b=0[x+2=0\Rightarrow x=-2]$

$\Rightarrow -2a+b=8\to (1)$

Again, given that,

$x-3$ is a factor of $x^3+ax+b$

$x-3$ is a factor of $x^3+ax+b$

$\Rightarrow$$(3{)}^3+a(3)+b=0$   $[x-3=0\Rightarrow x=3]$

$\Rightarrow$$3a+b=-27\to (2)$

On solving equations $(1)$ and $(2)$, we get $a=-7$ and $b=-6$.

11. Polynomial $x^3-a{x}^2+bx-6$ leaves remainder $-8$ when divided by $x-1$ and $x-2$ is a factor of it. Find the values of ' $a$ ' and ' $b$ '.

Ans:

On dividing by $x-1$, the polynomial $x^3-b{x}^2+bx-6$ leaves the remainder $-8$.

$\Rightarrow (1{)}^3-a(1{)}^2+b(1)-6=-8[x-1=0\Rightarrow x=1]$

$\Rightarrow$$-a+b=-3$

$\Rightarrow a-b=3\to (1)$

$(x-2)$ is a factor of polynomial $x^3-b{x}^2+bx-6$.

$\Rightarrow (2{)}^3-a(2{)}^2+b(2)-6=0$

$\Rightarrow 8-4a+2b-6=0$

$\Rightarrow 2a-b=1\to (2)$

When subtracting $(1)$ and $(2)$, we get $a=-2$.

When substituting the value of $a$ in $(1)$, we get  $b=-54$.

12. Using the Factor Theorem, show that $(x-2)$ is a factor of $3x^2-5x-2$ Hence, factorise the given expression.

Ans:

Since, $x-2=0\Rightarrow x=2$

$\therefore$ Remainder $=$ The value of $3x^2-5x-2$ at $x=2$.

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

$=12-10-2=0$

$\Rightarrow (x-2)$ is a factor of $3x^2-5x-2$.

Now, divide $\left(3x^2-5x-2\right)$ by $(x-2)$,

$$\begin{gathered} 3x+1 \\ x-2)\overline{3x^2-5x-2} \\ \underset{―}{3x^2-6x} \\ x-2 \\ \underset{―}{x-2} \\ 0 \end{gathered}$$

The quotient $=3x+1$.

Therefore, $3x^2-5x-2=(x-2)(3x+1)$.

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

Ans:

$2x+7=0\Rightarrow x=-\frac{7}{2}$

Remainder $=\text{ Value of }2x^3+5x^2-11x-14\text{ at }x=-\frac{7}{2}$

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

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

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

$=0$

$\Rightarrow (2x+7)\text{ is a factor of }2x^3+5x^2-11x-14$

Now, divide $2x^3+5x^2-11x-14$ by $2x+7$,

$$\begin{gathered} x^2-x-2 \\ 2x+7)\overline{2x^3+5x^2-11x-14} \\ \underset{―}{2x^3+7x^2} \\ -2x^2-11x-14 \\ \underset{―}{-2x^2-7x} \\ -14x-14 \\ \underset{―}{-14x-14} \\ 0 \end{gathered}$$

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

$=(2x+7)\left(x^2-2x+x-2\right)$

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

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

14. Using the Remainder Theorem, factorise the expression $2x^3+x^2-2x-1$ completely.

Ans:

For $x=1$, the value of given expression is,

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

$=2+1-2-1=0$

$\Rightarrow x-1$ is a factor of $2x^3+x^2-2x-1$.

Now, divide $2x^3+x^2-2x-1$ by $x-1$,

$$\begin{gathered} 2x^2+3x+1 \\ x-1)\overline{2x^3+x^2-2x-1} \\ \underset{―}{2x^3-2x^2} \\ 3x^2-2x-1 \\ \underset{―}{3x^2-3x} \\ x-1 \\ \underset{―}{x-1} \\ 0 \end{gathered}$$

Thus, $2x^3+x^2-2x-1=(x-1)\left(2x^2+3x+1\right)$

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

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

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

15. Find the values of ' $a$ ' and ' $b$ ' so that the polynomial $x^3+a{x}^2+bx-45$ has $(x-1)$ and $(x+5)$ as its factors. For the values of ' $a$ ' and ' $b$ ', as obtained above, factorise the given polynomial completely.

Ans:

$(x-1)$ is a factor of given polynomial $x^3+a{x}^2+bx-45$

$\Rightarrow$$(1{)}^3+a(1{)}^2+b(1)-45=0$  $[x-1=0\Rightarrow x=1]$

$\Rightarrow a+b=44\to (1)$

$(x+5)$ is a factor of given polynomial,

$\Rightarrow (-5{)}^3+a(-5{)}^2+b(-5)-45=0[x+5=0\Rightarrow x=-5]$

$\Rightarrow -125+25a-5b-45=0$

$\Rightarrow 5a-b=34\to (2)$

When adding $(1)$ and $(2)$, we get $a=13$.

When substituting the value of $a$ in $(1)$, we get $b=31.$

$\therefore$ The given polynomial $x^3+a{x}^2+bx-45$ $=x^3+13x^2+31x-45$

Now divide this polynomial by $(x-1)$,

$$\begin{gathered} x^2+14x+45 \\ x-1)\overline{x^3+13x^2+31x-45} \\ \underset{―}{x^3-x^2} \\ 14x^2+31x-45 \\ \underset{―}{14x^2-14x} \\ 45x-45 \\ \underset{―}{45x-45} \\ 0 \end{gathered}$$

Thus, $x^3+13x^2+31x-45$

$=(x-1)\left(x^2+14x+45\right)$

$=(x-1)\left(x^2+9x+5x+45\right)$

$=(x-1)[x(x+9)+5(x+9)]$

$=(x-1)(x+9)(x+5)$

16. If $(x-2)$ is a factor of $2x^3-x^2-px-2$

(i) find the value of $p$.

Ans:

$x-2=0\Rightarrow x=2$

Since, $(x-2)$ is a factor of given expression

 $\therefore$ Remainder $=0$

$\Rightarrow 2(2{)}^3-(2{)}^2-p\times 2-2=0$

$\Rightarrow 10-2p=0$

$\Rightarrow p=5$

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

Ans:

Substitute the value of $p$in given polynomial,

$2x^3-x^2-px-2=2x^3-x^2-5x-2$

Now, divide $2x^3-x^2-5x-2$ by $x-2$,

$$\begin{gathered} 2x^2+3x+1 \\ x-2)\overline{2x^3-x^2-5x-2} \\ \underset{―}{2x^3-4x^2} \\ 3x^2-5x-2 \\ \underset{―}{3x^2-6x} \\ x-2 \\ \underset{―}{x-2} \\ 0 \end{gathered}$$

Thus, $2x^3-x^2-5x-2$

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

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

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

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

EXERCISE 8(A)

1. Find, in each case, the remainder when:

(i) $x^4-3x^2+2x+1$ is divided by $x-1$.

Ans:

By the remainder theorem we know that when a polynomial $f(x)$ is divided by $x-a$, then the remainder is $f(a)$.

Here, $f(x)=x^4-3x^2+2x+1$

Remainder$=f(1)=(1{)}^4-3(1{)}^2+2(1)+1$

$=1-3+2+1$

$=1$

(ii) $x^3+3x^2-12x+4$ is divided by $x-2$.

Ans:

By the remainder theorem we know that when a polynomial $f(x)$ is divided by $x-a$, then the remainder is $f(a)$.

Here, $f(x)=x^3+3x^2-12x+4$

Remainder$=f(2)=(2{)}^3+3(2{)}^2-12(2)+4$

$=8+12-24+4$

$=0$

(iii) $x^4+1$ is divided by $x+1$.

Ans:

By the remainder theorem we know that when a polynomial $f(x)$ is divided by $x-a$, then the remainder is $f(a)$.

Here, $f(x)=x^4+1$

Remainder$=f(-1)=(-1{)}^4+1$

$=1+1$

$=2$

2. Show that:

(i) $x-2$ is a factor of $5x^2+15x-50$

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Then, $x-2=0$

$x=2$

Remainder$=f(2)$

$=5(2{)}^2+15(2)-50$

$=0$

The remainder of the polynomial $f(x)=5x^2+15x-50$  is $0$ .

Therefore,  $x-2$  is a factor of polynomial $f(x)=5x^2+15x-50$ .

(ii) $3x+2$ is a factor of $3x^2-x-2$.

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Then, $3x+2=0$

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

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

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

$=0$

The remainder of the polynomial $3x^2-x-2$  is $0$ .

Therefore,  $3x+2$  is a factor of polynomial $3x^2-x-2$ .

3. Use the Remainder Theorem to find which of the following is a factor of $2x^3+3x^2-5x-6$.

(i) $x+1$

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Then, $x+1=0$

$x=-1$

Remainder$=f(-1)$

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

$=0$

The remainder of the polynomial $f(x)=2x^3+3x^2-5x-6$ is $0$ .

Therefore, $x+1$ is a factor of polynomial $f(x)=2x^3+3x^2-5x-6$.

(ii) $2x-1$

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Then, $2x-1=0$

$x=\frac{1}{2}$

Remainder $=f\left(\frac{1}{2}\right)$ 

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

$=-\frac{15}{2}$

The remainder of the polynomial $f(x)=2x^3+3x^2-5x-6$ is $-\frac{15}{2}\ne 0.$

Therefore, $2x-1$ is a factor of polynomials $f(x)=2x^3+3x^2-5x-6$.

(iii) $x+2$

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Then, Then, $x+2=0$

$x=-2$

Remainder$=f(-2)$

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

$=0$

The remainder of the polynomial $f(x)=2x^3+3x^2-5x-6$ is $0$.

Therefore, $x+2$ is a factor of polynomials $f(x)=2x^3+3x^2-5x-6$.

4. (i) If $2x+1$ is a factor of $2x^2+ax-3$, find the value of $a$.

Ans:

Here, $2x+1$ is a factor of $f(x)=2x^2+ax-3$.

Then, $f\left(\frac{-1}{2}\right)=0$

$\Rightarrow 2{\left(\frac{-1}{2}\right)}^2+a\left(\frac{-1}{2}\right)-3=0$

$\Rightarrow \frac{1}{2}-\frac{a}{2}=3$

$\Rightarrow 1-a=6$

$\Rightarrow a=-5$

(ii) Find the value of $k$, if $3x-4$ is a factor of expression $3x^2+2x-k$.

Ans:

Here, $3x-4$ is a factor of $f(x)=3x^2+2x-k$.

Then, $f\left(\frac{4}{3}\right)=0$

$\Rightarrow 3{\left(\frac{4}{3}\right)}^2+2\left(\frac{4}{3}\right)-k=0$

$\Rightarrow \frac{16}{3}+\frac{8}{3}-k=0$

$\Rightarrow \frac{24}{3}=k$

$\Rightarrow k=8$

5. Find the values of constants $a$ and $b$ when $x-2$ and $x+3$ both are the factors of expression $x^3+a{x}^2+bx-12$.

Ans:

Let $f(x)=x^3+a{x}^2+bx-12$ 

$x-2$ is a factor of $f(x)$. 

Then, $x=2$

Now, remainder $=0$

$(2{)}^3+a(2{)}^2+b(2)-12=0$

$\Rightarrow 8+4a+2b-12=0$

$\Rightarrow 4a+2b-4=0$

$\Rightarrow 2a+b-2=0\to (1)$

$x+3$ is a factor of $f(x)$.

Then, $x=-3$

Now, remainder $=0$

$(-3{)}^3+a(-3{)}^2+b(-3)-12=0$

$\Rightarrow -27+9a-3b-12=0$

$\Rightarrow 9a-3b-39=0$

$\Rightarrow 3a-b-13=0\to (2)$

Adding $$\left(1\right)$$ and$\left(2\right)$, we get,

$5a-15=0$

$\Rightarrow a=3$

Putting the value of a in $$\left(1\right)$$, we get,

$6+b-2=0$

$\Rightarrow b=-4$

Thus, the value of constant $a$ is $3$ and $b$is $-4.$

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

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Let $f(x)=(3k+2)x^3+(k-1)$

$2x+1=0\Rightarrow x=\frac{-1}{2}$

Since, $2x+1$ is a factor of $f(x)$, remainder is $0$.

Then, $(3k+2){\left(\frac{-1}{2}\right)}^3+(k-1)=0$

$\Rightarrow \frac{-(3k+2)}{8}+(k-1)=0$

$\Rightarrow \frac{-3k-2+8k-8}{8}=0$

$\Rightarrow 5k-10=0$

$\Rightarrow k=2$

7. Find the value of $a$, if $x-2$ is a factor of $2x^5-6x^4-2a{x}^3+6a{x}^2+4ax+8$.

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Let$f(x)=2x^5-6x^4-2a{x}^3+6a{x}^2+4ax+8$ 

$x-2=0\Rightarrow x=2$ 

Since, $x-2$ is a factor of $f(x)$, remainder $=0$ 

Then, $2(2{)}^5-6(2{)}^4-2a(2{)}^3+6a(2{)}^2+4a(2)+8=0$ 

$64-96-16a+24a+8a+8=0$ 

$-24+16a=0$ 

$16a=24$ 

$a=1.5$

8. Find the values of $m$ and $n$ so that $x-1$ and $x+2$ both are factors of $x^3+\left(3m+1\right)x^2+nx-18.$

Ans:

When a polynomial $f(x)$ is divided by $x-a$, the remainder $=f(a)$. And if the remainder $f(a)=0;x-a$ is a factor of the polynomial $f(x)$.

Let $f(x)=x^3+(3m+1)x^2+nx-18$ 

$x-1=0\Rightarrow x=1$ 

$x-1$ is a factor of $f(x)$. 

So, remainder $=0$ 

Now, $(1{)}^3+(3m+1)(1{)}^2+n(1)-18=0$