factorise x^3+12x^2+39x+28
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Answered by
3
Answer:
x3+12x2+39x+28
=x3+x2+11x2+11x+28x+28
=x2(x+1)+11x(x+1)+28(x+1)
=(x+1)(x2+11x+28)
=(x+1)(x2+7x+4x+28)
=(x+1)[x(x+7)+4(x+7)]
=(x+1)(x+4)(x+7)
Step-by-step explanation:
not sure if it's right ☺️
Answered by
4
Answer:
p(x)=x³+12x²+39x+28
factors of 28=±1,±2,±7
p(-1)=(-1)³+12(-1)²+39(-1)+28
=-1+12-39+28
=-40+40
=0
so,x+1 is a factor of x³+12x²+39x+28
x³+12x²+39x+28=(x+1)(x²+11x+28)
=(x+1)(x²+4x+7x+28)
=(x+1)[x(x+4)+7(x+4)]
=(x+1)(x+4)(x+7)
hope this helps you
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