Find the zeroes of the following polynomials by factorisation

(i) $4 x^{2}-3 x-1$

Solution:

Let   $f(x)=4 x^{2}-3 x-1$

$=4 x^{2}-4 x+x-1 \quad$ [by splitting the middle term]

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

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

So, the value of $4 x^{2}-3 x-1$ is zero when $x-1=0$ or $4 x+1=0$ i.e., when $x=1$ or $x=-\frac{1}{4}$.

So, the zeroes of $4 x^{2}-3 x-1$ are 1 and $-\frac{1}{4}$.

$\therefore$ Sum of zeroes $=1-\frac{1}{4}=\frac{3}{4}=\frac{-(-3)}{4}$

$=(-1)\left(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\right)$

and product of zeroes $=(1)\left(-\frac{1}{4}\right)=-\frac{1}{4}$

$=(-1)^{2} \cdot\left(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\right)$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(ii) $3 x^{2}+4 x-4$

Solution:

Let $\quad f(x)=3 x^{2}+4 x-4$

$=3 x^{2}+6 x-2 x-4 \quad$ [by splitting the middle term]

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

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

So, the value of $3 x^{2}+4 x-4$ is zero when $x+2=0$ or $3 x-2=0$, i.e., when $x=-2$ or

$x=\frac{2}{3} .$ So, the zeroes of $3 x^{2}+4 x-4$ are $-2$ and $\frac{2}{3} .$

$\therefore \quad$ Sum of zeroes $=-2+\frac{2}{3}=-\frac{4}{3}$

$=(-1) \cdot\left(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\right)$

and $\quad$ product of zeroes $=(-2)\left(\frac{2}{3}\right)=\frac{-4}{3}$

$=(-1)^{2} \cdot\left(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\right)$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(iii) $51^{2}+12 t+7$.

Solution:

Let $\quad f(t)=5 t^{2}+12 t+7$

$=5 t^{2}+7 t+5 t+7$    [by splitting the middle term]

$=t(5 t+7)+1(5 t+7)$

$=(5 t+7)(t+1)$

So, the value of $5 t^{2}+12 t+7$ is zero when $5 t+7=0$ or $t+1=0$,

i.e., when $t=\frac{-7}{5}$ or $t=-1$

So, the zeroes of $5 t^{2}+12 t+7$ are $-7 / 5$ and $-1$

$\therefore$ Sum of zeroes $=-\frac{7}{5}-1=\frac{-12}{5}$

$=(-1) \cdot\left(\frac{\text { Coefficient of } t}{\text { Coefficient of } t^{2}}\right)$

and product of zeroes $=\left(-\frac{7}{5}\right)(-1)=\frac{7}{5}$

$=(-1)^{2} \cdot\left(\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\right)$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(iv) $t^{3}-2 t^{2}-15 t$

Solution:

Let

$f(t)=t^{3}-2 t^{2}-15 t$

$=t\left(t^{2}-2 t-15\right)$

$=t\left(t^{2}-5 t+3 t-15\right)$   [by splitting the middle term]

$=t[t(t-5)+3(t-5)]$

$=t(t-5)(t+3)$

So, the value of $t^{3}-2 t^{2}-15 t$ is zero when $t=0$ or $t-5=0$ or $t+3=0$

i.e., $\quad$ when $t=0$ or $t=5$ or $t=-3$.

So, the zeroes of $t^{3}-2 t^{2}-15 t$ are $-3,0$ and 5 .

$\therefore \quad$ Sum of zeroes $=-3+0+5=2=\frac{-(-2)}{1}$

$=(-1) \cdot\left(\frac{\text { Coefficient of } t^{2}}{\text { Coefficient of } t^{3}}\right)$

Sum of product of two zeroes at a time

$=(-3)(0)+(0)(5)+(5)(-3)$

$=0+0-15=-15$

$=(-1)^{2} \cdot\left(\frac{\text { Coefficient of } t}{\text { Coefficient of } t^{3}}\right)$

and product of zeroes $=(-3)(0)(5)=0$

$=(-1)^{3}\left(\frac{\text { Constant term }}{\text { Coefficient of } t^{3}}\right)$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(v) $2 x^{2}+\frac{7}{2} x+\frac{3}{4}$

Solution:

Let $\quad f(x)=2 x^{2}+\frac{7}{2} x+\frac{3}{4}=8 x^{2}+14 x+3$

$=8 x^{2}+12 x+2 x+3$    [by splitting the middle term]

$=4 x(2 x+3)+1(2 x+3)$

$=(2 x+3)(4 x+1)$

So, the value of $8 x^{2}+14 x+3$ is zero when $2 x+3=0$ or $4 x+1=0$,

i.e., when $x=-\frac{3}{2}$ or $x=-\frac{1}{4}$.

So, the zeroes of $8 x^{2}+14 x+3$ are $-\frac{3}{2}$ and $-\frac{1}{4}$

$\therefore \quad$ Sum of zeroes $=-\frac{3}{2}-\frac{1}{4}=-\frac{7}{4}=\frac{-7}{2 \times 2}$

$=-\frac{\text { (Coefficient of } x \text { ) }}{\text { (Coefficient of } x^{2} \text { ) }}$

And $\quad$ roduct of zeroes $=\left(-\frac{3}{2}\right)\left(-\frac{1}{4}\right)=\frac{3}{8}=\frac{3}{2 \times 4}$

$=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(vi) $4 \times 2+5 \sqrt{2 x}-3$

Solution:

Let $f(x)=4 x^{2}+5 \sqrt{2} x-3$

$=4 x^{2}+6 \sqrt{2} x-\sqrt{2} x-3$   [by splitting the middle term]

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

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

So, the value of $4 x^{2}+5 \sqrt{2} x-3$ is zero when $\sqrt{2} x+3=0$ or $2 \sqrt{2} \cdot x-1=0$,

i.e., when $x=-\frac{3}{\sqrt{2}}$ or $x=\frac{1}{2 \sqrt{2}}$

So, the zeroes of $4 x^{2}+5 \sqrt{2} x-3$ are $-\frac{3}{\sqrt{2}}$ and $\frac{1}{2 \sqrt{2}}$.

$\therefore$ Sum of zeroes $=-\frac{3}{\sqrt{2}}+\frac{1}{2 \sqrt{2}}$

$=-\frac{5}{2 \sqrt{2}}=\frac{-5 \sqrt{2}}{4}$

$=-\frac{\text { (Coefficient of } x \text { ) }}{\text { (Coefficient of } x^{2} \text { ) }}$

and product of zeroes $=-\frac{3}{\sqrt{2}} \cdot \frac{1}{2 \sqrt{2}}=-\frac{3}{4}$ 

$=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(vii) $2 s^{2}-(1+2 \sqrt{2}) s+\sqrt{2}$

Solution:

Let $\quad f(s)=2 s^{2}-(1+2 \sqrt{2}) s+\sqrt{2}$

$=2 s^{2}-s-2 \sqrt{2} s+\sqrt{2}$

$=s(2 s-1)-\sqrt{2}(2 s-1)$

$=(2 s-1)(s-\sqrt{2})$

So, the value of $2 s^{2}-(1+2 \sqrt{2}) s+\sqrt{2}$ is zero when $2 s-1=0$ or $s-\sqrt{2}=0$,

i.e., $\quad$ when $s=\frac{1}{2}$ or $s=\sqrt{2}$.

So, the zeroes of $2 s^{2}-(1+2 \sqrt{2}) s+\sqrt{2}$ are $\frac{1}{2}$ and $\sqrt{2}$.

$\therefore \quad$ Sum of zeroes $=\frac{1}{2}+\sqrt{2}=\frac{1+2 \sqrt{2}}{2}=\frac{-[-(1+2 \sqrt{2})]}{2}=\frac{(\text { Coefficient of } s)}{\left(\text { Coefficient of } s^{2}\right)}$

and product of zeroes $=\frac{1}{2} \cdot \sqrt{2}=\frac{1}{\sqrt{2}}=\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(viii) $v^{2}+4 \sqrt{3 v}-15$

Solution:

Let $\quad f(v)=v^{2}+4 \sqrt{3} v-15$

$=v^{2}+(5 \sqrt{3}-\sqrt{3}) v-15$

$=v^{2}+5 \sqrt{3} v-\sqrt{3} v-15$  [by splitting the middle term]

$=v(v+5 \sqrt{3})-\sqrt{3}(v+5 \sqrt{3})$

$=(v+5 \sqrt{3})(v-\sqrt{3})$

So, the value of $v^{2}+4 \sqrt{3} v-15$ is zero when $v+5 \sqrt{3}=0$ or $v-\sqrt{3}=0$,

i.e., $\quad$ when $v=-5 \sqrt{3}$ or $v=\sqrt{3}$.

So, the zeroes of $v^{2}+4 \sqrt{3} v-15$ are $-5 \sqrt{3}$ and $\sqrt{3}$.

$\therefore \quad$ Sum of zeroes $=-5 \sqrt{3}+\sqrt{3}=-4 \sqrt{3}$

$=(-1) \cdot\left(\frac{\text { Coefficient of } v}{\text { Coefficient of } v^{2}}\right)$

and $\quad$ product of zeroes $=(-5 \sqrt{3})(\sqrt{3})$

$=-5 \times 3=-15$

$=(-1)^{2} \cdot\left(\frac{\text { Constant term }}{\text { Coefficient of } v^{2}}\right)$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(ix) $y^{2}+\frac{3}{2} \sqrt{5 y}-5$

Solution:

Let $\quad f(y)=y^{2}+\frac{3}{2} \sqrt{5} y-5$

$=2 y^{2}+3 \sqrt{5} y-10$

$=2 y^{2}+4 \sqrt{5} y-\sqrt{5} y-10$   [by splitting the middle term]

$=2 y(y+2 \sqrt{5})-\sqrt{5}(y+2 \sqrt{5})$

$=(y+2 \sqrt{5})(2 y-\sqrt{5})$

So, the value of $y^{2}+\frac{3}{2} \sqrt{5} y-5$ is zero when $(y+2 \sqrt{5})=0$ or $(2 y-\sqrt{5})=0$,

i.e., $\quad$ when $y=-2 \sqrt{5}$ or $y=\frac{\sqrt{5}}{2}$

So, the zeroes of $2 y^{2}+3 \sqrt{5} y-10$ are $-2 \sqrt{5}$ and $\frac{\sqrt{5}}{2}$

$\therefore \quad$ Sum of zeroes $=-2 \sqrt{5}+\frac{\sqrt{5}}{2}=\frac{-3 \sqrt{5}}{2}=-\frac{\text { (Coefficient of } y \text { ) }}{\text { (Coefficient of } y^{2} \text { ) }}$

And product of zeroes $=-2 \sqrt{5} \times \frac{\sqrt{5}}{2}=-5=\frac{\text { Constant term }}{\text { Coefficient of } y^{2}}$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.

(x) $7 y^{2}-\frac{11}{3} y-\frac{2}{3}$.

Solution:

Let $f(y)=7 y^{2}-\frac{11}{3} y-\frac{2}{3}$

$=21 y^{2}-11 y-2$

$=21 y^{2}-14 y+3 y-2$  [by splitting the middle term]

$=7 y(3 y-2)+1(3 y-2)$

$=(3 y-2)(7 y+1)$

So, the value of $7 y^{2}-\frac{11}{3} y-\frac{2}{3}$ is zero when $3 y-2=0$ or $7 y+1=0$,

i.e., $\quad$ when $y=\frac{2}{3}$ or $y=-\frac{1}{7}$.

So, the zeroes of $7 y^{2}-\frac{11}{3} y-\frac{2}{3}$ are $\frac{2}{3}$ and $-\frac{1}{7}$.

$\therefore \quad$ Sum of zeroes $=\frac{2}{3}-\frac{1}{7}=\frac{14-3}{21}=\frac{11}{21}=-\left(\frac{-11}{3 \times 7}\right)$

$=(-1) \cdot\left(\frac{\text { Coefficient of } y}{\text { Coefficient of } y^{2}}\right)$

and product of zeroes $=\left(\frac{2}{3}\right)\left(-\frac{1}{7}\right)=\frac{-2}{21}=\frac{-2}{3 \times 7}$

$=(-1)^{2} \cdot\left(\frac{\text { Constant term }}{\text { Coefficient of } y^{2}}\right)$

Hence, verified the relations between the zeroes and the coefficients of the polynomial.