Question

Factorise x3+7x2+7x-15 by factor theorem

Answer 1

Given :  x³ + 7x² + 7x - 15

To Find : Factorise

Solution:

x³ + 7x² + 7x - 15

x = 1

=  1³ + 7(1)² + 7(1) - 15

= 1 + 7 + 7 - 15

= 15 - 15

= 0

Hence x - 1 is one of the factor

               x² + 8x + 15

            _______________

x - 1   _|  x³ + 7x² + 7x - 15  |_

              x³ - x²

           _____________

                     8x²  + 7x  - 15  

                     8x²  - 8x

                  ______________

                              15x  - 15

                              15x  - 15

                       ____________

                                   0

                           _______

x³ + 7x² + 7x - 15  = (x - 1)(x² + 8x + 15)

x² + 8x + 15

= x² + 5x + 3x + 15

= x(x + 5) + 3(x + 5)

= (x + 5)(x + 3)

x³ + 7x² + 7x - 15   = (x-1)(x + 5)(x + 3)

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Answer 2

$\underline{\textsf{Given:}}$

$\mathsf{Polynomial\;is\;x^3+7x^2+7x-15}$

$\underline{\textsf{To find:}}$

$\mathsf{Factors\;of\;x^3+7x^2+7x-15}$

$\underline{\textsf{Solution:}}$

$\textsf{Factor theorem:}$

$\boxed{\mathsf{(x-a)\;is\;a\;factor\;P(x)\;\iff\;P(a)=0}}$

$\mathsf{Let\;P(x)=x^3+7x^2+7x-15}$

$\mathsf{Sum\;of\;the\;coefficients=1+7+7-15=0}$

$\therefore\mathsf{(x-1)\;is\;a\;factor\;of\;P(x)}$

$\mathsf{When\;x=-3}$

$\mathsf{P(-3)=(-3)^3+7(-3)^2+7(-3)-15}$

$\mathsf{P(-3)=-27+63-21-15}$

$\mathsf{P(-3)=63-63}$

$\mathsf{P(-3)=0}$

$\therefore\mathsf{(x+3)\;is\;a\;factor}$

$\mathsf{When\;x=-5}$

$\mathsf{P(-5)=(-5)^3+7(-5)^2+7(-5)-15}$

$\mathsf{P(-5)=-125+175-35-15}$

$\mathsf{P(-5)=175-175}$

$\mathsf{P(-5)=0}$

$\therefore\mathsf{(x+5)\;is\;a\;factor}$

$\underline{\textsf{Answer:}}$

$\mathsf{x^3+7x^2+7x-15=(x-1)(x+3)(x+5)}$

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