How do you factor x^2-8x-15 ?

2 Answers
Sep 25, 2016

Use the Completing the Square Method to get ((x+4)+sqrt(31))((x+4)-sqrt(31)).

Explanation:

The Completing the Square Method basically "forces" the existence of a perfect square trinomial in order to easily factor an equation where factoring by grouping is impossible.

For x^2-8x-15, the first step would be to make the equation equal to zero and add 15 to both sides.

x^2-8x=15

Next, we need to turn the binomial on the left side of the equation into a perfect trinomial. We can do this by dividing the coefficient of the "middle x-term", which would be -8, in half then squaring it.

(-8)/2=-4 and (-4)^2=16

We then add the result to both sides of the equation.

x^2-8x+16=15+16

Now, we can factor the perfect square trinomial and simplify 15+16.

(x+4)^2=31

Now, we subtract 31 from both sides of the equation.

(x+4)^2-31

In order to factor these terms, both of them need to be the "squared version of their square rooted form" so that the terms stay the same when factoring.

(x+4)^2-sqrt(31)^2

Now, we can factor them. Since the expression follows the case a^2-b^2, where, in this case, a=(x+4) and b=sqrt(31), we can factor them by following the Difference of Squares Formula: a^2-b^2=(a+b)(a-b).

Your final answer would be:
((x+4)+sqrt(31))((x+4)-sqrt(31))

Sep 25, 2016

x^2-8x-15=(x-4-sqrt31)(x-4+sqrt31)

Explanation:

In the quadratic polynomial ax^2+bx+c, where a, b and c are rational numbers, if b^2-4ac is positive but not the square of a rational number, roots are irrational, in such cases it may not be possible to factorize the polynomial. In such cases factorization, we may use quadratic formula (-b+--sqrt(b^2-4ac))/(2a) to find zeros, and if they are p and q, the factors are a(x-p)(x-q).

In x^2-8x-15, b^2-4ac=(-8)^2-4xx1xx(-15)=64+60=124 is not the square of a rational number, and using quadratic formula roots are

(-(-8)+--sqrt((-8)^2-4xx1xx(-15)))/(2xx1)

= (8+-sqrt124)/2=(8+-2sqrt31)/2=4+-sqrt31, and factors are

x^2-8x-15=(x-4-sqrt31)(x-4+sqrt31)