Show that (x-2) and (x+3) are factor of x^3 -6x^2 -13x+42
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Step-by-step explanation:
If (x-2) and (x+3) are factor of p(x)= x^3-6x^2-13x+42 , then
P(2)=0 & p(-3)=0
=> p(2)= (2)^3-6(2)^2-13×2+42
= 8-6×4-26+42
= 8-24-26+42 = - 50 + 50 = 0
=> p(-3)= (-3)^3-6(-3)^2-13×(-3)+42
= -27-6×9+39+42
= -27-54+39+42 = -81+81 = 0
=> 2 and -3 are zeroes of p(x)
=> (x-2) and (x+3) are factors of p(x)
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