one of the factor of polynomial x^3+4x^2-3x-18 is x+3 then factorise the given polynomial
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x^3 +4x^2 - 3x - 18
= x^3 + ( 3x^2 + x^2) + ( 3x - 6x) - 18
= x^3 + 3x^2 + x^2 +3x - 6x - 18
= x^2 ( x + 3) + x (x +3) - 6(x +3)
= (x +3)[ x^2 + x - 6]
= (x +3)[ x^2 + 3x - 2x - 6]
= (x +3) [ x (x +3) - 2(x +3)]
= (x +3)( x +3)(x - 2)
= x^3 + ( 3x^2 + x^2) + ( 3x - 6x) - 18
= x^3 + 3x^2 + x^2 +3x - 6x - 18
= x^2 ( x + 3) + x (x +3) - 6(x +3)
= (x +3)[ x^2 + x - 6]
= (x +3)[ x^2 + 3x - 2x - 6]
= (x +3) [ x (x +3) - 2(x +3)]
= (x +3)( x +3)(x - 2)
Answered by
0
x^3 +4x^2 - 3x - 18
= x^3 + ( 3x^2 + x^2) + ( 3x - 6x) - 18
= x^3 + 3x^2 + x^2 +3x - 6x - 18
= x^2 ( x + 3) + x (x +3) - 6(x +3)
= (x +3)[ x^2 + x - 6]
= (x +3)[ x^2 + 3x - 2x - 6]
= (x +3) [ x (x +3) - 2(x +3)]
= (x +3)( x +3)(x - 2)
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