Question

How do you factor examples b), e) and f) below?

b) `y = 8x^2-10x-3`

e) `y = 5x^3-80x`

f) `y = 36x^2-1`

Answer 1

b) `y = 8x^2-10x-3 = (4x+1)(2x-3)`

e) `y = 5x^3-80x = 5x(x-4)(x+4)`

f) `y = 36x^2-1 = (6x-1)(6x+1)`

Explanation:

Example b)

`y = 8x^2-10x-3`

This quadratic is in the form `ax^2+bx+c`, with `a=8`, `b=-10` and `c = -3`.

In order to tell whether it factors 'nicely', let us first check the discriminant:

`Delta = b^2-4ac = (-10)^2-4(8)(-3) = 100+96 = 196 = 14^2`

Since this is a perfect square, the given quadratic will factor exactly.

Let's use an AC method to find the factors:

Look for a pair of factors of `AC = 8*3 = 24` which differ by `10`.

The pair `12, 2` works.

Use this pair to split the middle term and factor by grouping:

`y = 8x^2-10x-3`

`color(white)(y) = 8x^2-12x+2x-3`

`color(white)(y) = (8x^2-12x)+(2x-3)`

`color(white)(y) = 4x(2x-3)+1(2x-3)`

`color(white)(y) = (4x+1)(2x-3)`

`color(white)()`
Example e)

`y = 5x^3-80x`

For this example, we can separate out the common factor `5x`, then use the difference of squares identity:

`a^2-b^2 = (a-b)(a+b)`

with `a=x` and `b=4` as follows:

`y = 5x^3-80x`

`color(white)(y) = 5x(x^2-16)`

`color(white)(y) = 5x(x^2-4^2)`

`color(white)(y) = 5x(x-4)(x+4)`

`color(white)()`
Example f)

`y = 36x^2-1`

Notice that both the terms are perfect squares, so we can use the difference of squares identity again...

`y = 36x^2-1 = (6x)^2-1^2 = (6x-1)(6x+1)`