Polynomials : Exercise - 2.4 Optional (Mathematics NCERT Class 10th)

Q.1        Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :
(i) $2{x^3} + {x^2} - 5x + 2\,;{1 \over 2},1, - 2$
(ii) ${x^3} - 4{x^2} + 5x - 2\,;2,1,1$

Sol.      (i) Comparing the given polynomial with $a{x^3} + b{x^2} + cx + d,$ we get
a = 2 , b = 1, c = – 5 and d = 2.
$p\left( {{1 \over 2}} \right) = 2{\left( {{1 \over 2}} \right)^3} + {\left( {{1 \over 2}} \right)^2} - 5\left( {{1 \over 2}} \right) + 2$
$= {1 \over 4} + {1 \over 4} - {5 \over 2} + 2$
$= {{1 + 1 - 10 + 8} \over 4} = {0 \over 4} = 0$
$p\left( 1 \right) = 2{\left( 1 \right)^3} + {\left( 1 \right)^2} - 5\left( 1 \right) + 2$
$= 2 + 1 - 5 + 2 = 0$
$p\left( { - 2} \right) = 2{\left( { - 2} \right)^3} + {\left( { - 2} \right)^2} - 5\left( { - 2} \right) + 2$
$= 2\left( { - 8} \right) + 4 + 10 + 2$
$= - 16 + 16 = 0$
Therefore, ${1 \over 2}$, 1 and – 2 are the zeroes of $2{x^3} + {x^2} - 5x + 2$.
So, $\alpha = {1 \over 2},\,\beta = 1\,and\,\gamma = - 2$.
Therefore, $\alpha + \beta + \gamma = {1 \over 2} + 1+\left( { - 2} \right) = {{1 + 2 - 4} \over 2}$
$= - {1 \over 2} = {{ - b} \over a}$
$\alpha \beta + \beta \gamma + \gamma \alpha = \left( {{1 \over 2}} \right)\left( 1 \right) + \left( 1 \right)\left( { - 2} \right) + \left( { - 2} \right)\left( {{1 \over 2}} \right)$
$= {1 \over 2} - 2 - 1$
$= {{1 - 4 - 2} \over 2} = {{ - 5} \over 2} = {c \over a}$
and     $\alpha \beta \gamma = {1 \over 2} \times 1 \times - 2 = 1 = {{ - 2} \over 2} = {{ - d} \over \alpha }$

(ii) Comparing the given polynomial with $a{x^3} + b{x^2} + cx + d$, we get
a = 1, b = – 4, c = 5 and d = – 2.
$p\left( 2 \right) = {\left( 2 \right)^3} - 4{\left( 2 \right)^2} + 5\left( 2 \right) - 2 = 8 - 16 + 10 - 2 = 0$
$p\left( 1 \right) = {\left( 1 \right)^3} - 4{\left( 1 \right)^2} + 5\left( 1 \right) - 2 = 1 - 4 + 5 - 2 = 0$
Therefore , 2 , 1 and 1 are the zeros of  ${x^3} - 4{x^2} + 5x - 2$
Thus,     $\alpha = 2,\,\beta = 1\,and\,\gamma = 1$.
Now $\alpha + \beta + \gamma = 2 + 1 + 1 = 4 = {{ - \left( { - 4} \right)} \over 1} = {{ - b} \over a}$
$\alpha \beta + \beta \gamma + \gamma \alpha = \left( 2 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 2 \right)$
$= 2 + 1 + 2 = 5 = {5 \over 1} = {c \over a}$
and           $\alpha \beta \gamma = \left( 2 \right)\left( 1 \right)\left( 1 \right) = 2 = {{ - \left( { - 2} \right)} \over 1} = {{ - d} \over a}$

Q.2       Find a cubic polynomial with the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.
Sol.     Let zeroes be $\alpha ,\beta \,\,and\,\,\gamma$
We have given ${\alpha + \beta + \gamma = 2}$
${\alpha \beta + \beta \gamma + \gamma \alpha = - 7}$
and $\alpha \beta \gamma = - 14$
Cubic polynomial is given by $\left( {x - \alpha } \right)\left( {x - \beta } \right)\left( {x - \,\gamma } \right) = 0$
$\Rightarrow {x^3} - {x^2}\left( {\alpha + \beta + \gamma } \right) + x\left( {\alpha \beta + \beta \gamma + \gamma \alpha } \right) - \alpha \beta \gamma = 0$
$\Rightarrow {x^3} - 2{x^2} - 7x + 14 = 0$
$\alpha \beta \gamma = - 14 = {{ - 14} \over 1} = {{ - d} \over a}$
If a = 1, then b = – 2, c = – 7 and d = 14
So, one cubic polynomial which fits the given conditions will be ${x^3} - 2{x^2} - 7x + 14$.

Q.3       If the zeroes of the polynomial ${x^3} - 3{x^2} + x + 1$ are a – b, a, a + b , find a and b
Sol.       Since (a – b) , a (a + b) are the zeroes of the polynomial ${x^3} - 3{x^2} + x + 1,$
Therefore, $\left( {a - b} \right) + a + \left( {a + b} \right) = {{ - \left( { - 3} \right)} \over 1} = 3$
$\Rightarrow$ 3a = 3 $\Rightarrow$ a = 1
(a – b) a + a(a + b) + (a + b) (a – b) $= {1 \over 1} = 1$
$\Rightarrow$ ${a^2} - ab + {a^2} + ab + {a^2} - {b^2} = 1$
$\Rightarrow$ $3{a^2} - {b^2} = 1$
$\Rightarrow$ $3{\left( 1 \right)^2} - {b^2} = 1$ [Since a = 1]
$\Rightarrow$ $3 - {b^2} = 1$
$\Rightarrow$ ${b^2} = 2$
$\Rightarrow$ $b = \pm \sqrt 2$
Hence , a = 1 and $b = \pm \sqrt 2$.

Q.4       If two zeroes of the polynomial ${x^4} - 6{x^3} - 26{x^2} + 138x - 35{\mkern 1mu} \,\,are\,\,{\mkern 1mu} 2 \pm \sqrt 3$ , find other zeroes.
Sol.       Since $\,{\mkern 1mu} 2 \pm \sqrt 3$, are two zeroes of the polynomial $p\left( x \right) = {x^4} - 6{x^3} - 26{x^2} + 138x - 35$
Let $x = 2 \pm \sqrt 3$ $\Rightarrow$ $x - 2 = \pm \sqrt 3$
Squaring we get ${x^2} - 4x + 4 = 3$
$\Rightarrow$ ${x^2} - 4x + 1 = 0$
Let us divide p(x) by ${x^2} - 4x + 1$ to obtain other zeroes. Therefore, $p\left( x \right) = {x^4} - 6{x^3} - 26{x^2} + 138x - 35$
$= \left( {{x^2} - 4x + 1} \right)\left( {{x^2} - 2x - 35} \right)$
$= \left( {{x^2} - 4x + 1} \right)\left( {{x^2} - 7x + 5x - 35} \right)$
$= \left( {{x^2} - 4x + 1} \right)\left[ {x\left( {x - 7} \right) + 5\left( {x - 7} \right)} \right]$
$= \left( {{x^2} - 4x + 1} \right)\left( {x + 5} \right)\left( {x - 7} \right)$
$\Rightarrow$ (x + 5) and (x – 7) are other factors of p(x).
Therefore,  – 5 and 7 are other zeroes of the given polynomial.

Q.5       If the polynomial ${x^4} - 6{x^3} + 16{x^2} - 25x + 10$ is divided by another polynomial ${x^2} - 2x + k$, the remainder comes out to be x + a find k and a.
Sol.       Let us divide ${x^4} - 6{x^3} + 16{x^2} - 25x + 10$ by ${x^2} - 2x + k$ So, remainder = (2 k – 9) x – (8 – k) k + 10
But the remainder is given as x + a.
On comparing their coefficients , we have  2 k – 9 = 1
$\Rightarrow$ 2 k = 10
$\Rightarrow$ k = 5
and                             – (8 – k)k + 10 = a
$\Rightarrow$ a = – (8 – 5)5 + 10  = – 3 × 5 + 10 = – 15 + 10 = – 5
Hence, k = 5 and a = – 5.