Question

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

Answer 1

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))`

Answer 2

`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)`