Polynomial – Important Question and Answers

    Question 1: Find the coefficient of $x^{2}$ in $\left(3x^{2}+2x-4\right)\left(x^{2}-3x-2\right)$

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

    $=3x^{4}-9x^{3}-6x^{2}+2x^{3}-6x^{2}-4x-4x^{2}+12x+8$

    Terms containing $x^{2}$ are : $-6x^{2}-6x^{2}-4x^{2}=-16x^{2}$

    So, coefficient of $x^{2}$ is -16.


    Question 2: For the polynomial $\frac{x^{3}+2x+1}{5}-\frac{7}{2}x^{2}-x^{6}$ , write :

    (i) the degree of the polynomial
    (ii) the coefficient of $x^{3}$
    (iii) the coefficient of $x^{6}$
    (iv) the constant term

    Answer: (i) The highest power of the variable is 6. So, the degree of the polynomial is 6.

    (ii) Coefficient of $x^{3}$ is $\frac{1}{5}$

    (iii) Coefficient of $x^{6}$ -1

    (iv) Constant term is $\frac{1}{5}$


    Question 3: Find a zero of the polynomial $p(x) = 2x + 1$.

    Answer : Finding a zero of $p(x)$, is the same as solving the equation $p(x) = 0$

    Now, $2x + 1 = 0$ gives us $x=-\frac{1}{2}$

    So, $x=-\frac{1}{2}$ is a zero of the polynomial $2x + 1$.


    Question 4: Verify that $1$ is not a zero of the polynomial $4y^{4}-3y^{3}+2y^{2}-5y+1$ .

    Answer : Let $p(y) = 4y^{4}-3y^{3}+2y^{2}-5y+1$

    Now, $p(1) = 4 – 3 + 2 – 5 + 1 = – 1 ≠ 0$
    ∴ $1$ is not a zero of $p(y)$.


    Question 5: Find the remainder when $x^{3}-ax^{2}+6x-a$ is divided by $x-a$

    Answer: $p(x)=x^{3}-ax^{2}+6x-a$

    When $p(x)$ is divided by $x – a$, the remainder is $p(a)$.
    Substitute $x = a$ in $p(x)$

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


    Question 6: If $p(x) = q(x) × g(x) + r(x), r(x) ≠ 0$,where $p(x), q(x), g(x),$ and $r(x)$ are polynomials,then what can we say about the degree of $r(x)$?

    Answers: Degree of remainder is always less than that of the divisor.


    Question 7: Find the remainder when the polynomial $f(x)=4x^{3}-12x^{2}+14x-3$ is divided by $(2x – 1)$.

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

    By the remainder theorem, we know that when $f (x)$ is divided by $(2x – 1)$, the remainder is $f\frac{1}{2}$

    $f\left(\frac{1}{2}\right)=[4\times\left(\frac{1}{2}\right)^{3}-12\times\left(\frac{1}{2}\right)^{2}+14\times\frac{1}{2}-3]=\left[\frac{1}{2}-3+7-3\right]=\frac{3}{2}$

    Hence, the required remainder is $\frac{3}{2}$


    Question 8: Check whether $7 + 3x$ is a factor of $3x^{3}+7x$ .

    Answer : $7+3x=0\Rightarrow3x=-7\Rightarrow x=\frac{-7}{3}$

    Substitute , $x=\frac{-7}{3}$ in $p(x)=3x^{3}+7x$

    $p\left(\frac{-7}{3}\right)=3(\frac{-7}{3})^{3}+7\left(\frac{-7}{3}\right)$

    $=\frac{343}{9}-\frac{49}{3}=\frac{-343-147}{9}=\frac{-490}{9}$

    So, remainder = $\frac{-490}{9}$

    Therefore, $(3x + 7)$ is not a factor of the polynomial $3x^{3}+7x$


    Question 9: The polynomial $ax^{3}+3x^{2}-3$ and $2x^{3}-5x+a$ a leave the same remainder in each case when divided by $(x – 4)$. Find the value of $a$.

    Answer: As the remainder is same, therefore,

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

    $\Rightarrow64a+48-3=128-20+a$

    $\Rightarrow64a+45=108+a$

    $\Rightarrow63a = 108 – 45 = 63$

    $\Rightarrow a = 1$


    Question 10 : If the polynomial $p(x)=x^{4}-2 x^{3}+3 x^{2}$ $-a x+8$ is divided by $(x-2),$ it leaves $a$ remainder $10 .$ Find the value of a.

    Answer: When $p(x)$ is divided by $x-2$ the remainder is $10 $
    $\therefore p(2)=(2)^{4}-2(2)^{3}+3(2)^{2}-a(2)+8=10 \Rightarrow 16-16+12-2 a+8=10 \Rightarrow-2 a=-10 \Rightarrow a=5$


    Question 11: The polynomials $x^{3}+2 x^{2}-5 a x-8$ and $x^{3}+a x^{2}-12 x-6$ when divided by $(x-2)$ and $(x-3)$ leave remainder p and q respectively. If $q-p=10,$ find the value of $a$.

    Answer:

    Let $p(x)=x^{3}+2 x^{2}-5 a x-8$
    $\therefore p(2)=(2)^{3}+2(2)^{2}-5 a(2)-8=p$
    $\Rightarrow 8+8-10 a-8=p$
    $\Rightarrow 8-10 a=p$ …….(i)

    $\text { Now, let } Q(x)=x^{3}+a x^{2}-12 x-6 $
    $\therefore Q(3)=(3)^{3}+a(3)^{2}-12(3)-6=q $
    $\quad=27+9 a-36-6=q $
    $\Rightarrow-15+9 a=q $ ………(ii)
    On subtracting (i) from (ii), we get $q-p$
    $\Rightarrow-15+9 a-8+10 a=10[\because q-p=10] $
    $\Rightarrow-23+19 a=10 \Rightarrow a=\frac{33}{19}$



    Question 12: The polynomials $a x^{3}+3 x^{2}-13$ and $2 x^{3}-5 x+a$ are divided by $x+2 .$ If the remainder in each case is the same, find the value of a.

    Answer: Let $p(x)=a x^{3}+3 x^{2}-13$ and $q(x)=2 x^{3}-5 x+a$ be the given polynomials. The remainders when $p(x)$ and $q(x)$ are divided by $(x+2)$ are $p(-2)$ and $q(-2)$ respectively.

    By the given condition, we have $ p(-2) &=q(-2) \Rightarrow & a(-2)^{3}+3 \times(-2)^{2}-13 \ &=2 \times(-2)^{3}-5 \times-2+a $

    $\Rightarrow &-8 a+12-13=-16+10+a $
    $\Rightarrow-8 a-1=a-6 \Rightarrow-9 a=-5 \Rightarrow a=\frac{5}{9}$


    Question 13: Factorise:
    (i) $12 x^{2}-7 x+1$
    (ii) $2 x^{2}+7 x+3$
    (i i i) $6 x^{2}+5 x-6$
    (iv) $3 x^{2}-x-4$
    Answer:
    (i) $12 x^{2}-7 x+1$ $=12 x^{2}-4 x-3 x+1$
    $=4 x(3 x-1)-1(3 x-1)$
    $=(4 x-1)(3 x-1)$
    (ii) $2 x^{2}+7 x+3$ $=2 x^{2}+6 x+x+3$
    $=2 x(x+3)+1(x+3)$
    $=(2 x+1)(x+3)$
    (iii) $6 x^{2}+5 x-6$ $=6 x^{2}+9 x-4 x-6$
    $=3 x(2 x+3)-2(2 x+3)$
    $=(3 x-2)(2 x+3)$
    (iv) $3 x^{2}-x-4$ $=3 x^{2}-4 x+3 x-4$
    $=x(3 x-4)+1(3 x-4)$
    $=(x+1)(3 x-4)$


    Question 14: If $(y-p)$ is a factor of $y^{6}-p y^{5}+y^{4}$ $-p y^{3}+3 y-p-2,$ then find the value of $p$

    Answer: $(y-p)$ is a factor of the polynomial

    $P(y) =y^{6}-p y^{5}+y^{4}-p y^{3}+3 y-p-2 $ ,then

    $ P(p)=0$

    $\Rightarrow p^{6}-p^{6}+p^{4}-p^{4}+3 p-p-2=0 $

    $\Rightarrow 2 p-2=0 \Rightarrow p=1 $


    Question 15: Find the values of $a$ and $b$ in which $x^{2}-1$ is a factor of $x^{4}+a x^{3}-3 x^{2}+2 x+b$

    Answer: Let $p(x)=x^{4}+a x^{3}-3 x^{2}+2 x+b$

    If $x^{2}-1=(x-1)(x+1)$ i.e., $(x-1)$ and $(x+1)$ are its factors, then each of $p(1)$ and $p(-1)$ is equal to $0$

    Now, $p(1)=(1)^{4}+a(1)^{3}-3(1)^{2}+2(1)+b=0$
    $=1+a-3+2+b=0$
    $=a+b=0$ ……(i)
    $=(-1)^{4}+a(-1)^{3}-3(-1)^{2}+2(-1)+b=0$
    $=1-a-3-2+b=0$
    $=-a-4+b=0$
    $=-a+b=4$ ……..(ii)
    Subtracting
    (i) from (ii), we get $a=-2$ and $b=2$


    Question 16. Factorise $: 2 x^{3}-5 x^{2}-19 x+42$

    Answer. Let $p(x)=2 x^{3}-5 x^{2}-19 x+42$
    Now, factors of 42 are 1,2,3,6,7,14,21,42
    $\therefore p(1) =2(1)^{3}-5 \cdot(1)^{2}-19+42 \\
    =2-5-19+42=20 \neq 0$
    $\therefore(x-1)$ is not a factor of $p(x)$

    Now, $p(2)=2(2)^{3}-5(2)^{2}-19(2)+42$
    $=16-20-38+42=0$
    $\therefore(x-2)$ is a factor of $p(x)$ Dividing $p(x)$ by $x-2,$ we get the other factors.
    $p(x) =(x-2)\left(2 x^{2}-x-21\right) $
    $=(x-2)\left(2 x^{2}-7 x+6 x-21\right) $
    $=(x-2)[x(2 x-7)+3(2 x-7)] $
    $=(x-2)(2 x-7)(x+3)$


    Question 17: Check whether $g(x)$ is a factor of $p(x)$ or not, where :
    $p(x)=8 x^{3}-6 x^{2}-4 x+3, g(x)=\frac{x}{3}-\frac{1}{4}$

    Answer: $g(x)=\frac{x}{3}-\frac{1}{4}=0$ gives $x=\frac{3}{4} \cdot g(x)$ will be a factor of $p(x)$ if $p\left(\frac{3}{4}\right)=0$ (Factor theorem)

    Now, $p\left(\frac{3}{4}\right)=8\left(\frac{3}{4}\right)^{3}-6\left(\frac{3}{4}\right)^{2}-4\left(\frac{3}{4}\right)+3$
    $=8 \times \frac{27}{64}-6 \times \frac{9}{16}-3+3=0$
    since, $p\left(\frac{3}{4}\right)=0,$ so, $g(x)$ is a factor of $p(x)$


    Question 18: Without actual division prove that $x^{4}+2 x^{3}-2 x^{2}+2 x-3$ is exactly divisible by $x^{2}+2 x-3$

    Answer: Let $p(x)=x^{4}+2 x^{3}-2 x^{2}+2 x-3$ and $q(x)=x^{2}+2 x-3=(x+3)(x-1)$

    $p(x)$ is exactly divisible by $q(x)$ means, $q(x)$ is a factor of $p(x)$
    $\Rightarrow(x+3)(x-1)$ is a factor of $p(x)$

    $\Rightarrow p(-3)=0$ and $p(1)=0$

    Now, $p(-3)$ $=(-3)^{4}+2(-3)^{3}-2(-3)^{2}+2(-3)-3$

    $=81-54-18-6-3=81=81-81=0$

    Again, $p(1)=1^{4}+2 \times 1^{3}-2 \times 1^{2}+2 \times 1-3$

    $=1+2-2+2-3=5-5=0$

    since, $p(-3)=0$ and $p(-1)=0,$ so $(x+3)$ and $(x-1)$ is a factor of $p(x)$ $\Rightarrow(x+3)(x-1)=x^{2}+2 x-3$ is a factor of $p(x)$

    $\Rightarrow p(x)$ is exactly divisible by $x^{2}+2 x-3$


    Question 19: If $a x^{3}+b x^{2}+x-6$ has $(x+2)$ as a factor and leaves a remainder 4 when divided by $x-2,$ find the value of a and b.

    Answer: Let $p(x)=a x^{3}+b x^{2}+x-6$ be the given polynomial.

    Now, $(x+2)$ is a factor of $p(x)$. $\Rightarrow p(-2)=0 \quad[As, x+2=0 \Rightarrow x=-2]$

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

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

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

    $\Rightarrow-2 a+b=2$ ………..(i)
    It is given that $p(x)$ leaves the remainder 4 when it is divided by $(x-2)$.

    $\therefore p(2)=4 \quad[\because x-2=0 \Rightarrow x=2]$

    $\Rightarrow a(2)^{3}+b(2)^{2}+2-6=4$

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

    $\Rightarrow 8 a+4 b=8 \Rightarrow 2 a+b=2 \quad \ldots$ ……….(ii)

    Adding (i) and (ii), we get $2 b=4 \Rightarrow b=2$

    Putting $b=2$ in (i), we get $-2 a+2=2 \Rightarrow-2 a=0 \Rightarrow a=0$

    Hence, $a=0$ and $b=2$


    Question 20. Factorise each of the following :
    (i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
    (ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$
    (iii) $27-125 a^{3}-135 a+225 a^{2}$
    (i v) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$

    Answer:

    (i) $8 a^{3}+b^{3}+12 a^{2} b+6 a b^{2}$
    $=(2 a)^{3}+b^{3}+3(2 a)(b)(2 a+b)$
    $=(2 a+b)^{3}=(2 a+b)(2 a+b)(2 a+b)$

    (ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}$
    $=(2 a)^{3}+(-b)^{3}+3(2 a)(-b)(2 a-b)$
    $=(2 a-b)^{3}=(2 a-b)(2 a-b)(2 a-b)$

    (iii) $27-125 a^{3}-135 a+225 a^{2}$
    $=3^{3}+(-5 a)^{3}+3 \times(3)(-5 a)(3-5 a)$
    $=(3-5 a)^{3}=(3-5 a)(3-5 a)(3-5 a)$

    (iv) $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$
    $=(4 a)^{3}+(-3 b)^{3}+3(4 a) \times(-3 b)(4 a-3 b)$
    $=(4 a-3 b)^{3}=(4 a-3 b)(4 a-3 b)(4 a-3 b)$


    Question 21: Find the value of$ (x-y)^{3}+(y-z)^{3}+(z-x)^{3}$ .

    Answer: $a^{3}+b^{3}+c^{3}=3 a b c $
    if $a+b+c=0 $
    Here, $ x-y+y-z+z-x=0$
    $\therefore(x-y)^{3}+(y-z)^{3}+(z-x)^{3} $
    $=3(x-y)(y-z)(z-x)$


    Question 22. If $a b=5$ and $a-b=2$, then find the value of $a^{3}-b^{3}$.

    Answer. $a-b=2 \Rightarrow a^{2}+b^{2}-2 a b=4$
    $\Rightarrow a^{2}+b^{2}-10=4 \Rightarrow a^{2}+b^{2}=14$

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

    $=2(14+5)=2 \times 19=38 $


    Question 23: Simplify $(x+y)^{3}-\left(x^{3}+y^{3}\right)$

    Answer. $(x+y)^{3}-\left(x^{3}+y^{3}\right)$
    $=x^{3}+y^{3}+3 x y(x+y)-x^{3}-y^{3}$
    $=3 x y(x+y)$


    Question 24: Factorise :
    $2 \sqrt{2} a^{3}+8 b^{3}-27 c^{3}+18 \sqrt{2} a b c$

    Answer. $2 \sqrt{2} a^{3}+8 b^{3}-27 c^{3}+18 \sqrt{2} a b c$
    $=(\sqrt{2} a)^{3}+(2 b)^{3}+(-3 c)^{3}-3 \times \sqrt{2} a \times 2 b$
    $=(\sqrt{2} a+2 b-3 c)\left[(\sqrt{2} a)^{2}+(2 b)^{2}+(-3 c)^{2}\right]$
    $-\sqrt{2} a \times 2 b-2 b \times(-3 c)-\sqrt{2} a \times(-3 c)$
    $=(\sqrt{2} a+2 b-3 c)\left(2 a^{2}+4 b^{2}+9 c^{2}-2 \sqrt{2} a b\right)+6 b c+3 \sqrt{2} a c)$


    Question 25: Simplify :
    $\frac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}$

    Answer. We have, $\left(a^{2}-b^{2}\right)+\left(b^{2}-c^{2}\right)+\left(c^{2}-a^{2}\right)=0$
    $\therefore\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}$
    $=3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)$
    $=3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$
    Similarly, we have,
    $(a-b)+(b-c)+(c-a)=0$
    $\Rightarrow(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$
    $=3(a-b)(b-c)(c-a)$
    $\therefore \frac{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}{(a-b)^{3}+(b-c)^{3}+(c-a)^{3}}$

    $=\frac{3(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{3(a-b)(b-c)(c-a)}$
    $=(a+b)(b+c)(c+a)$