Question

How do you factor `20x^2 + 37x + 15`?

Answer 1

Use the quadratic formula to find the roots of `20x^2+37x+15=0` and hence that:

`20x^2+37x+15 = (4x+5)(5x+3)`

Explanation:

Let `f(x) = 20x^2+37x+15`

This is of the form `ax^2+bx+c`, with `a=20`, `b=37` and `c=15`.

This has determinant `Delta` given by the formula:

`Delta = b^2-4ac = 37^2 - (4xx20xx15) = 1369-1200 = 169 = 13^2`

Since this is a perfect square, `f(x) = 0` has rational roots and `f(x)` has factors with rational coefficients.

Since we have gone to the trouble of computing the determinant, we might as well use the quadratic formula to find the roots of `f(x) = 0` as:

`x = (-b+-sqrt(b^2-4ac))/(2a) = (-b+-sqrt(Delta))/(2a) = (-37+-13)/40`

That is: `x = (-37-13)/40 = -50/40 = -5/4`

or: `x = (-37+13)/40 = -24/40 = -3/5`

Hence `f(x) = (4x+5)(5x+3)`

Answer 2

Use a version of the AC Method to find:

`20x^2+37x+15=(4x+5)(5x+3)`

Explanation:

Let `A=20`, `B=37`, `C=15`.

Look for a factorization of `AC=20*15=300=2^2*3*5^2` into a pair of factors `B1` and `B2` whose sum is `B=37`.

Note that one of `B1` and `B2` must be divisible by `25` since otherwise both `B1` and `B2` would be divisible by `5` and so would their sum, but `37` is not divisible by `5`.

Try `B1=25`. Then `B2=300/(B1)=300/25=12`

That works: `25+12 = 37`.

Next, for each of the pairs `(A, B1)` and `(A, B2)` divide by the HCF (highest common factor) to get a pair of coefficients of a factor of our original quadratic...

`(A, B1) = (20, 25) -> (4, 5) -> (4x+5)`

`(A, B2) = (20, 12) -> (5, 3) -> (5x+3)`

So `20x^2+37x+15=(4x+5)(5x+3)`

Answer 3

Factor y = 20x^2 + 37x + 15

Ans: (5x + 3)(4x + 5)

Explanation:

`y = 20x^2 + 37x + 15 =` 20(x - p)(x - q)

I use the new AC Method to factor trinomials.
Converted trinomial `y' = x^2 + 37 x + 300 =` (x - p')(x - q').
Factor pairs of (300) -> ...(6, 50)(10, 30)(12, 25). This sum is 35 = b.
Then, p' = 12, and q' = 25.
Therefor, `p = (p')/a = 12/20 = 3/5,` and `q = (q')/a = 25/20 = 5/4.`
Factored form: `y = 20(x + 3/5)(x + 5/4) = (5x + 3)(4x + 5)`