Question

How do you factor `3x ^ { 2} + 8x - 528= 0`?

Answer 1

`x=12` or `x=-44/3`

Explanation:

To factor `ax^2+bx+c=0`, we should split the middle term `b` in two parts, so that the product of two parts is `ac`.

In `3x^2+8x-528=0`, we should split the middle term `8` in two parts, so that the product of two parts is `3xx-528=1584`.

As product is negative, the factors of `-1584` will be opposite in sign and as their sum is `8`, the difference in absolute value is `8` and positive factor is greater.

Factors of `1584` are `(2,792),(3,528),(4,396),(6,264),(8,198),(9,176),(11,144),(12,132),(16,99),(18,88),(22,72),(24,66),(33,48),(36,44)`

Now `36` and `44` have a difference of `8`

Hence, `3x^2+8x-528=0` can bewritten as

`3x^2+44x-36x-528=0`

or `x(3x+44)-12(3x+44)=0`

or `(x-12)(3x+44)=0`

i.e. `x=12` or `x=-44/3`

Answer 2

`(x-12)(3x+44) = 0`

Explanation:

This can be factored by completing the square.

Given:

`f(x) = 3x^2+8x-528`

To make the arithmetic less messy, we can premultiply by `3` to make the leading coefficient into a perfect square, then divide by `3` at the end.

We will also use the difference of squares identity:

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

with `a=(3x+4)` and `b=40` as follows:

`3f(x) = 3(3x^2+8x-528)`

`color(white)(3f(x)) = 9x^2+24x-1584`

`color(white)(3f(x)) = (3x)^2+2(3x)(4)+16-1600`

`color(white)(3f(x)) = (3x+4)^2-40^2`

`color(white)(3f(x)) = ((3x+4)-40)((3x+4)+40)`

`color(white)(3f(x)) = (3x-36)(3x+44)`

`color(white)(3f(x)) = 3(x-12)(3x+44)`

Dividing both ends by `3` we find:

`3x^2+8x-528 = (x-12)(3x+44)`

So the given equation can be written:

`(x-12)(3x+44) = 0`

which has zeros `x=12` and `x=-44/3`

Answer 3

`(3x+44)(x-12)=0`

`x = -44/3 or x = 12`

Explanation:

The first step is to determine whether there are factors.

`a=3" "b = 8 and c =-528`

`b^2-4ac" "rarr 8^2 -4(3)(-528)=6400`

`6400` is a perfect square, therefore there are rational factors.

We need to find factors of `3 and 528` whose products differ by `8`.

The smaller the value of `b`, the closer it is to `sqrt(ac)`

`ac = 3xx528 = 1584`

`sqrt1584 = 39.799 ~~ 40`

Use some trial and error starting from `40`

`40` is not a factor of `1584`
`41` is not a factor of `1584`
`42` is not a factor of `1584`
`43` is not a factor of `1584`

`color(red)(44 xx 36 =1584 and 44-36 =8)" "larr` BINGO!

Now combine factors of `3 and 528` to get to `44 and 36`

`" "3 and 528`
`" "darrcolor(white)(xxxx)darr`
`" "3color(white)(xxxxx)44" "rarr 1xx44 = 44`
`" "1color(white)(xxxxx)12" "rarr 3xx12 = ul36`
`color(white)(xxxxxxxxxxxxxxxxxxxx)8`

We have the factors, now add in the signs to get `+8`

`" "3 and" " 528`
`" "darrcolor(white)(xxxxx)darr`
`" "3color(white)(xxxxx)+44" "rarr 1xx44 = +44`
`" "1color(white)(xxxxx)-12" "rarr 3xx12 = -ul(36)`
`color(white)(xxxxxxxxxxxxxxxxxxxxxxxx)+8`

`(3x+44)(x-12)=0`

If we use the factors to solve the equation we get:

`3x+44 = 0 " "rarr x = -44/3`

`x-12 =0 " "rarr x = 12`