factorise (x³+6x²+11x+6)
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Let p(x) = (x³+6x²+11x+6), constant term of p(x) is 6
Factor of 6 is +1, +2, +3 and +6
Now, p(-1) = (-1)^3 + 6×(-1)^2 + 11×(-1) + 6
= -1 + 6×1 + (-11) +6
= -1 +6 -11 +6
= 0
By trial we find p(-1) =0 , so (x+1) is a factor of p(x)
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