show that(x+3)is a factor of the polynomial x³-3x²-10x+24
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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)
Step-by-step explanation:
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