Question

Question 1 (iv)
Find the zeroes of the following polynomials by factorization method and verify the relations between the zeroes and the coefficient of the polynomials
(iv) $t^3–2t^2–15t$

Answer 1

iv) Let $f\left(t\right)=t^3–2t^2–15t+\mathrm{0.....}\left(constanttermiszero\right)$
$=t(t^2–2t–15)$
$=t(t^2–5t+3t-15)$
$=t\left[t\right(t–5)+3(t-5\left)\right]$
$=t(t–5)(t+3)$
So, the value of $t^3–2t^2–15t$ is Zero when t = 0 or t – 5 = 0 or t + 3 = 0
i.e., when t = 0 or t = 5 or t = - 3
so, the zeroes of $t^3–2t^2–15tare–3,0and5$
$\therefore$ sum of zeroes = $-3+0+5=2=\frac{-(-2)}{1}$
$=(-1)\left(\frac{Coefficentof{t}^2}{Coefficientof{t}^3}\right)$
Sum of product of two zeroes at a time
$=(-3)\left(0\right)+\left(0\right)\left(5\right)+\left(5\right)(-3)$
$=0+0–15=-15$
$=(-1{)}^2\left(\frac{Coefficientoft}{Coefficientof{t}^3}\right)$
and product of zeroes $(-3)\left(0\right)\left(5\right)=0$
$=(-1{)}^3\left(\frac{Constantterm}{Coeffidientof{t}^3}\right)$
Hence, verified the relations between the zeroes and the coefficients of the polynomial.