Question

if p(x)=x^3-4x^2+×+6 then show that p(3)=0 and hence factorize p(x)​

Answer 1

p(x) = $x^3 - 4 x^2 +x+6$

p(3) = $3^3 -4 × 3^2 + 3 + 6$

= 27 - 36 + 9

= 0

HOPE THIS ANSWER HELPS U....

Answer 2

Step-by-step explanation:

$p(x) = {x}^{3} - 4 {x}^{2} + x + 6$

$p(0) = {0}^{3} - 4 \times {0}^{2} + 0 + 6$

$p(0) = 6$