Question

verify that 2,-2 and -3/2 are the zeros of the cuubic polynomial p(x)=2x^3+3x^2-8x-12 find relation

Answer 1

Answer:

Step-by-step explanation:

JUST SUBSTITUTE ONE BY ONE THE NUMBERS : 2, -2, -3\2 IN POLYNOMIAL AND IF VALUE IS 0, THEN IT IS A ZERO OF POLYNOMIAL.

Answer 2

Answer:

Step-by-step explanation:

$p(x)=2x^3+3x^2-8x-12\\\\p(2)=2\times(2)^3+3\times(2)^2-8\times(2)-12\\\\=2\times8+3\times4-16-12\\\\=16+12-16-12\\\\=0$

$\therefore p(2)=0$

$p(x)=2x^3+3x^2-8x-12\\\\p(-2)=2\times(-2)^3+3\times(-2)^2-8\times(-2)-12\\\\=2\times(-8)+3\times4+16-12\\\\=-16+12+16-12\\\\=0$

$\therefore p(-2)=0$

$p(x)=2x^3+3x^2-8x-12\\\\p(-\frac{3}{2})=2\times(-\frac{3}{2})^3+3\times(-\frac{3}{2})^2-8\times(-\frac{3}{2})-12\\\\=2\times(-\frac{27}{8})+3\times\frac{9}{4}+12-12\\\\=-\frac{27}{4}+\frac{27}{4}+12-12\\\\=0$

$\therefore p(-\frac{3}{2})=0$