Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: 2x3+x2-5x+2; -1/2, 1, -2
Given, p(x) = 2x3+x2-5x+2
And zeroes for p(x) are = 1/2, 1, -2
∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2
= (1/4)+(1/4)-(5/2)+2 = 0
p(1) = 2(1)3+(1)2-5(1)+2 = 0
p(-2) = 2(-2)3+(-2)2-5(-2)+2 = 0
Hence, proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.
Now, comparing the given polynomial with general expression, we get;
∴ ax3+bx2+cx+d = 2x3+x2-5x+2
a=2, b=1, c= -5 and d = 2
As we know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;
α +β+γ = –b/a
αβ+βγ+γα = c/a
α βγ = – d/a.
Therefore, putting the values of zeroes of the polynomial,
α+β+γ = ½+1+(-2) = -1/2 = –b/a
αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2)
= -5/2 = c/a
α β γ = ½×1×(-2) = -2/2 = -d/a
Hence, the relationship between the zeroes and the coefficients are satisfied.