Question

Find the zeros of the polynomial x2- 20x - 96

Answer 1

Answer:

$\Large \text{Zeroes are 24 and - 4}$

Step-by-step explanation:

$\Large \text{Given p(x)= x^2 -20x-96}\\\\\\\Large \text{we have to find zeroes of p(x)}\\\\\\\Large \text{First factorise the p(x)}\\\\\\\Large \text{p(x)= x^2 -20x-96}\\\\\\\Large \text{By splitting mid term }\\\\\\\Large \text{we have to make ac=+-b}\\\\\\\Large \text{p(x)= x^2 -24x+4x-96}\\\\\\\Large \text{p(x)=x(x-24)+4(x-24)}$

$\Large \text{p(x)=(x-24)(x+4)}\\\\\\\Large \text{Zeroes of p(x) }\\\\\\\Large \text{x-24=0 or x+4=0}\\\\\\\Large \text{x=24 or x= - 4}\\\\\\\Large \text{Thus we get our zeroes are 24 and -4}$

Answer 2

Answer:

Step-by-step explanation:

X^2-24x+4x-96

X(x-24)+4(x-24)

(x+4)(x-24)=0

X=-4,24