Factorise :
x^3 + 6x^2 + 12x + 16
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Answered by
2
Step-by-step explanation:
Let p(x) = x^3 + 6x^2 + 12x + 16
=> x^3 + 6x^2 + 12x + 8 + 8
=> x^3 + 3*2x^2 + 3*2*2*x + 2^3 + 8
=> (x+2)^3 + 2^3
=> (x+2+2) [(x+2)^2 - 2(x+2) + 4]
=> (x+4) (x^2+4x+4-2x-4+4)
=> (x+4) (x^2+2x+4)
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Answered by
11
x^3 + 6x^2 + 12x + 16
= x^3 + 6x^2 + 12x + 8 + 8
= x^3 + 2x*3x + 2*2*3x+ 2^3 + 2^3
= (x+2x) + 2^3
= ( x + 2 + 2) [ (x+2)² + 2(x+2) + 4)]
= ( x + 4 ) ( x² + 2x + 4 )
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