Question

How do you factor the quadratic equation `12x²+13x-4`?

Answer 1

In general, polynomials of any power
`x^n+a_1x^(n-1)+...+a_(n-1)x+a_n`
can be factored into a product of linear polynomials
`(x-b_1)(x-b_2)...(x-b_n)`
using their roots `b_1,b_2,...,b_n`, that is values of an unknown `x` that make the polynomial to evaluate to zero:
`b_i^n+a_1b_i^(n-1)+...+a_(n-1)b_i+a_n=0` for any `i=1,2,...,n`

In particular, a quadratic polynomial
`x^2+a_1x+a_2`
can be factored into a product of two linear polynomials
`(x-b_1)(x-b_2)`,
where `b_1` and `b_2` are solutions of an equation
`x^2+a_1x+a_2=0`

If there is a coefficient not equal to `1` at the highest power of an unknown `x`, we have to factor it out prior to the above transformations.

So, the straight forward method to factor this quadratic polynomial is to factor out `12` and to find two roots of the remaining polynomial, that is two solutions to an equation
`x^2+(13/12)x-4/12=0`

We can use a formula for solutions of this equation:
`x_(1,2)=(-13/12+-sqrt(13^2/12^2+4*4/12))/2=(-13+-19)/24`
From this we have two solutions:
`x_1=1/4` and `x_2=-4/3`

That brings the following factorization:
`12x^2+13x-4=12(x-1/4)(x+4/3)`

Answer 2

A quadratic expression is completely factorizable if and only if its discriminant is positive. Given a quadratic expression of the form `ax^2+bx+c`, the discriminant `\Delta` is defined as `b^2-4ac`. In your case, we have `a=12`, `b=13` and `c=-4`. For this values, we have `\Delta=361`, which means that we can factor the expression finding two solutions `x_1` and `x_2`, and thus writing `12x^2+13x-4=(x-x_1)(x-x_2)`.

To find the solutions, we have the formula
`x_{1,2}=\frac{-b\pm \sqrt(\Delta)}{2a}`.
Since `\Delta=361`, its square root equals 19. Plugging the values, we have the two solutions
`x_1={-13+19}/{24}=1/4` and
`x_2={-13-19}/{24}=-4/3`.

The factorization is thus `(x-1/4)(x+4/3)`

Answer 3

`12x^2 +13x-4 = (4x-1)(3x+4)`

Read though the method carefully.
It's worth the time spent to learn it.

Explanation:

`12x^2 +13x-4` is an example of a quadratic (`x^2`) trinomial.(3 terms)

Some trinomials are the product of two binomials of the form:

`(x+-a)(x+-b)`

When we try to find these factors from the quadratic, we are factoring or factorising. Not all quadratic trinomials can be factored.

In `color(magenta)(12)x^2 +color(lime)(13)x-color(magenta)(4)`

We are trying to find factors of 12 and 4 which subtract to make 13.
Look at the clues.

CLUE 1 : The sign with the 4 is negative

`rarr` The signs in the the brackets are different
`color(white)(xxxxxxx)(+) xx( -) rarr ( -)`
`rarr` The difference between the factors must be 13.

CLUE 2: 13 is an odd number, It can only be obtained from subtracting with an odd and an even number.

The factors of 4 are `1xx4 or cancel(2xx2)`
In this case it will not be `2xx2` because that will make both pairs even.

`rarr`The required factors of 4 are `1 and 4`

The factors of 12 are: `1xx12 and cancel(2xx6) and 3xx4`
`2 and 6` are both even - reject them as explained above.

Thinking through the possible combinations might go like this:

12x4 = 48 ....too big
12x1 = 12........too small - we still need to subtract.

Let's try 4 and 3 with 1 and 4. Cross-multiply and subtract

`color(white)(x.) (12)" "(4)`
`color(white)(x.x) darr" "darr`
`color(white)(xxx) 4" "1 rarr 3xx1 = 3`
`color(white)(xxx) 3" "4 rarr 4xx4 = 16 " subtract: " 16-3 = 13`

`rarr`We have the correct factors!! Now for the signs.

`+13 rarr` more positives. `color(red)(-3)color(blue)( +16) = +13`
Make them `color(red)(-3) and color(blue)( +16)`

`color(white)(xxxxxx)darr` Insert the signs here between the factors.

`color(white)(xxx) 4" "color(red)(-1)rarr 3xxcolor(red)(-1) = color(red)(-3)`
`color(white)(xxx) 3" "color(blue)(+4) rarr 4xxcolor(blue)(+4) = color(blue)(+16)`

The top row has the factors for the first bracket.
The bottom row has the factors for the second bracket.

`12x^2 +13x-4 = (4x-1)(3x+4)`

You can check by multiplying out again.

This requires a good deal of practice and a solid knowledge of the multiplication tables. However, once mastered it is a very quick method which works most of the time.

It does avoid having to find all the possible factors of numbers which are sometimes quite big.