Answer :
[tex]x^3+6x^2+5x-12=x^3-x^2+7x^2-7x+12x-12=\\\\=x^2(x-1)+7x(x-1)+12(x-1)=(x-1)(x^2+7x+12)=\\\\=(x-1)(x^2+3x+4x+12)=(x-1)[x(x+3)+4(x+3)]=\\\\=(x-1)(x+3)(x+4)[/tex]
x³ + 6x² + 5x - 12
= x³ - x² + 7x² + 12x - 12
If we add all the coefficients, we get that the answer is 0. Thus, (x-1) is a factor of polynomial.
= x²(x-1) + 7x(x-1) + 12(x-1)
on re-arranging (x-1) as a common factor ;
= (x-1)(x²+7x+12) .........................................(1)
Now, we factorize (x² + 7x + 12)
(x² + 7x + 12)
= x² + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 4)( x +3) ..........................................(2)
On substituting for p(x) in 1 and 2, we get
(px) = (x-1)(x+4)(x+3)
= x³ - x² + 7x² + 12x - 12
If we add all the coefficients, we get that the answer is 0. Thus, (x-1) is a factor of polynomial.
= x²(x-1) + 7x(x-1) + 12(x-1)
on re-arranging (x-1) as a common factor ;
= (x-1)(x²+7x+12) .........................................(1)
Now, we factorize (x² + 7x + 12)
(x² + 7x + 12)
= x² + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 4)( x +3) ..........................................(2)
On substituting for p(x) in 1 and 2, we get
(px) = (x-1)(x+4)(x+3)