How do you factor x squared - 5x - 6?

Answer 1

Normally, polynomials in this format can be factored by finding two numbers whose product is equal to the last term multiplied by the coefficient of the first term, and whose sum is equal to the coefficient of the second term. The polynomial can then be broken down further.

In this case though, there are no numbers that meet that condition, so this polynomial can not be factored into terms that use simple integers. In order to find it's factors then, what you'll need to do is set the value of the statement to zero, and solve for x. This will give you two different values, which will define the numeric terms in this polynomial's factors:

Let:

x2 + 6x - 4 = 0

∴ x2 + 6x + 9 = 13

∴ (x + 3)2 = 13

∴ x + 3 = ± √13

∴ x = -3 ± √13

So we have two possible values of x, the negatives of which will be our "a" and "b" values if we express the original polynomial in the format (x + a)(x + b):

[x - (-3 + √13)][x - (-3 - √13)]

= (x + 3 + √13)(x + 3 - √13)

Which you can expand in order to show that it is indeed equal:

= x2 + (3 - √13)x + (3 + √13)x + (3 - √13)(3 + √13)

= x2 + 6x + 9 + 3√13 - 3√13 - 13

= x2 + 6x + -4

Another way you can deal with this is to separate the last term into more than one value, using a technique called "completing the square". This allows you to reduce the rest of the polynomial:

x2 + 6x + -4

= x2 + 6x + 9 - 13

= (x + 3)2 - 13

Answer 2

If that's + 5x, the answer is (2x - 1)(3x + 4)

If that's - 5x, the answer is (2x + 1)(3x - 4)

Answer 3

Too bad that's not x^2 - 10x + 16

That factors to (x - 2)(x - 8)

Answer 4

You cannot. The roots of the quadratic are 5 ± sqrt(19). These roots are not rational and so there is no factorisation.

Answer 5

The expression does not have rational factors.

Answer 6

6x+4

factor:

2(3x+2)

Answer 7

6x2+x-35 = (2x+5)(3x-7)

Answer 8

(x - 6)(x + 1)

Answer 9

(2x + 5)(2x - 5)

Answer 10

(2x + 1)(3x - 4)