x^3 - 8x^2 + x +42 factorise using factor theorem

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To Factorise : x3-8x2+x+42 using Factor TheoremSolution : Since the contant term is 42So, factors of 42=±1, ±2, ±3, ±6, ±7, ±14, ±21, ±42let p(x)=x3-8x2+x+42So, putting x=-1 in p(x)p(-1)=(-1)3-8(-1)2+(-1)+42=-1-8-1+42=320So, x=-1 i.e. (x+1) is not a factor of x3-8x2+x+42Now,putting x=1 in p(x)p(1)=(1)3-8(1)2+(1)+42=1-8+1+42=36So, x=1 i.e. (x-1) is not a factor of x3-8x2+x+42Now,putting x=-2 in p(x)p(-2)=(-2)3-8(-2)2+(-2)+42=-8-32-2+42=0So, x=-1 i.e. (x+2) is  a factor of x3-8x2+x+42(x+2) will divide x3-8x2+x+42 completelyx2(x+2)-10x(x+2)+21(x+2)(x+2)(x2-10x+21)Let q(x)=x2-10x+21Since the contant term is 21So, factors of 21=±1, ±3, ±7, ±21 Since, we have already checked for x=±1 in x3-8x2+x+42So, now we shall check for x=-3putting x=-3 in q(x)q(-3)=(-3)2-10(-3)+21=9+30+21=600So, x=-3 i.e. (x+3) is not a factor of x2-10x+21Now,putting x=3 in q(x)q(3)=(3)2-10(3)+21=9-30+21=0So, x=3 i.e. (x-3) is a factor of x2-10x+21(x-3) will divide x2-10x+21 completelyx(x-3)-7(x-3)(x-3)(x-7)Hence, factors of x3-8x2+x+42 are (x+2), (x-3) and (x-7)
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But give a value from which we will factorise this polynomial
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