Question 18

Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials
(v) $2x^2+(\frac{\mathrm{7/}}{\mathrm{2)}}x+\frac{\mathrm{3/}}{4}$

Solution:

Let $f\left(x\right)=2x2+(7/2)x+3/4=8{\mathrm{x2}}^{}+14x+3$
$=8{\mathrm{x2}}^{}+12x+2x+3$ [by splitting the middle term]
$=4x(2x+3)+1(2x+3)$
$=(2x+3)(4x+1)$
So, the value of $8{\mathrm{x2}}^{}+14x=3$ is zero when $2x+3=0$ or $4x+1=0$.
i.e., when $x=-\frac{\mathrm{3/}}{2}$ or $x=-\frac{\mathrm{1/}}{4}$
So, the zeroes of $8{\mathrm{x2}}^{}+14x+3$ are $-\frac{\mathrm{3/}}{2}$ and $-\frac{\mathrm{1/}}{4}$
$\therefore$ Sum of zeroes $=-\frac{\mathrm{3/}}{2}-\frac{\mathrm{1/}}{4}=-\frac{\mathrm{7/}}{4}=\frac{-}{\mathrm{7/(}}$
$=-\frac{(}{\text{Coefficient of x}}$
And product of zeroes $=(-\frac{32}{})(-\frac{\mathrm{1/}}{4})=\frac{\mathrm{3/}}{8}=\frac{\mathrm{3/(}}{2}$
$=\frac{\text{Constant term/}}{\text{Coefficient of}}$
Hence, verified the relations between the zeroes and the coefficient of the polynomial.