Question

How do you factor `4x ^ { 2} - 29x + 51`?

Answer 1

`4x^2-29x+57 = (4x-17)(x-3)`

Explanation:

Some times determining the factors can be a bit challenging.

Consider the factors of 51

As soon as I observed that the last digit is 1 I immediately thought of `3xx7=21`

`51-21=30` short so lets stick a ten in there and see what happens. Lets try `(+-3)xx(+-17)=+51 larr" Works"`

So we have: `(?+-3)(?+-17)`

Notice that the `-29x` is negative. So perhaps we should instead test: `(?-3)(?-17)`

The `4` from `4x^2` is likely to have `1xx4" or "2xx2`

Lets test `2xx2`

`(2x-3)(2x-17) = 4x^2 color(red)(-34x larr" imediat fail"`

Lets test `4xx1`

`(x-3)(4x-17) = 4x^2-17x-12x+57`

`color(white)("ddddddddddddd") =4x^2-29x+57color(green)(larr" Works")`

`4x^2-29x+57 = (4x-17)(x-3)`

Answer 2

`4x^2-29x+51=(4x-17)(x-3)`

Explanation:

For factorizing `ax^2+bx+c` we find factors of `ac`, whose sum is `b`.

However, if signs of `a` and `c` are opposite take `ac` as positive and find factors of `ac`, whose difference is `b`.

Here we have `4x^2-29x+51` and our `ac` is `4xx51=204` and as their sign is same, we try to find factors whose sum is `29`.

Factors of `204` are `(2,102),(3,68),(4,51),(6,34),(12,17)`

Here `12` and `17` add up to `29`

Hence split middle term accordingly

`4x^2-29x+51`

= `4x^2-12x-17x+51`

= `4x(x-3)-17(x-3)`

= `(4x-17)(x-3)`

Additional Information: Had it been `4x^2-28x-51`, we would have found factors `(6,34)` whose product is `204` and difference is `28` (as sign of `4` and `51` are different, one being positive and other negative) and hence, we would have worked as follows:

`4x^2-28x-51`

= `4x^2+6x-34x-51`

= `2x(2x+3)-17(2x+3)`

= `(2x-17)(2x+3)`