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

Aug 5, 2015

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

$20 {x}^{2} + 37 x + 15 = \left(4 x + 5\right) \left(5 x + 3\right)$

#### Explanation:

Let $f \left(x\right) = 20 {x}^{2} + 37 x + 15$

This is of the form $a {x}^{2} + b x + c$, with $a = 20$, $b = 37$ and $c = 15$.

This has determinant $\Delta$ given by the formula:

$\Delta = {b}^{2} - 4 a c = {37}^{2} - \left(4 \times 20 \times 15\right) = 1369 - 1200 = 169 = {13}^{2}$

Since this is a perfect square, $f \left(x\right) = 0$ has rational roots and $f \left(x\right)$ 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 $f \left(x\right) = 0$ as:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 37 \pm 13}{40}$

That is: $x = \frac{- 37 - 13}{40} = - \frac{50}{40} = - \frac{5}{4}$

or: $x = \frac{- 37 + 13}{40} = - \frac{24}{40} = - \frac{3}{5}$

Hence $f \left(x\right) = \left(4 x + 5\right) \left(5 x + 3\right)$

Aug 5, 2015

Use a version of the AC Method to find:

$20 {x}^{2} + 37 x + 15 = \left(4 x + 5\right) \left(5 x + 3\right)$

#### Explanation:

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

Look for a factorization of $A C = 20 \cdot 15 = 300 = {2}^{2} \cdot 3 \cdot {5}^{2}$ into a pair of factors $B 1$ and $B 2$ whose sum is $B = 37$.

Note that one of $B 1$ and $B 2$ must be divisible by $25$ since otherwise both $B 1$ and $B 2$ would be divisible by $5$ and so would their sum, but $37$ is not divisible by $5$.

Try $B 1 = 25$. Then $B 2 = \frac{300}{B 1} = \frac{300}{25} = 12$

That works: $25 + 12 = 37$.

Next, for each of the pairs $\left(A , B 1\right)$ and $\left(A , B 2\right)$ divide by the HCF (highest common factor) to get a pair of coefficients of a factor of our original quadratic...

$\left(A , B 1\right) = \left(20 , 25\right) \to \left(4 , 5\right) \to \left(4 x + 5\right)$

$\left(A , B 2\right) = \left(20 , 12\right) \to \left(5 , 3\right) \to \left(5 x + 3\right)$

So $20 {x}^{2} + 37 x + 15 = \left(4 x + 5\right) \left(5 x + 3\right)$

Aug 5, 2015

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

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

#### Explanation:

$y = 20 {x}^{2} + 37 x + 15 =$ 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, $p = \frac{p '}{a} = \frac{12}{20} = \frac{3}{5} ,$ and $q = \frac{q '}{a} = \frac{25}{20} = \frac{5}{4.}$
Factored form: $y = 20 \left(x + \frac{3}{5}\right) \left(x + \frac{5}{4}\right) = \left(5 x + 3\right) \left(4 x + 5\right)$