Question

If α (alfa), β (bita), γ (gama) are the zeros of a cubic polynomial such that
α+β+γ = 3, αβ+βγ+αγ = -1 and αβγ = -24
then that polynomial is __________
(A). x³+3x²-10x-24
(B). x³+2x²-3x-24
(C). x³-3x²-10x+24
(D). x³-10x²-3x²-12

Answer 1

Answer:

Factors of x³-3x²-10x+24

Factors of x³-3x²-10x+24= (x-2)(x-4)(x+3)

Step-by-step explanation:

Let the polyomial p(x)=x³-3x²-10x+24

= x³+(-2x²-x²)+(2x-12x)+24

=(x³-2x²)+(-x²+2x)+(-12x+24)

=x²(x-2)-x(x-2)-12(x-2)

=(x-2)(x²-x-12)

=(x-2)(x²-4x+3x-12)

=(x-2)[x(x-4)+3(x-4)]

=(x-2)[(x-4)(x+3)]

=(x-2)(x-4)(x+3)

Therefore,

Factors of x³-3x²-10x+24

= (x-2)(x-4)(x+3)

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Answer 2

Let the polyomial p(x)=x³-3x²-10x+24

= x³+(-2x²-x²)+(2x-12x)+24

=(x³-2x²)+(-x²+2x)+(-12x+24)

=x²(x-2)-x(x-2)-12(x-2)

=(x-2)(x²-x-12)=(x-2)(x-4)(x+3)