# How do you factor 24x^2-76x+40?

May 15, 2015

First, let's simplify this dividing the equation by $4$.

$6 {x}^{2} - 19 x + 10$

Then, we have to find its roots: ${x}_{1} = \frac{2}{3}$ and ${x}_{2} = \frac{5}{2}$. But in order not to work with these fractions, we can multiply both sides by the denominator of the fractions, leaving only rational numbers:

$3 {x}_{1} = \frac{2}{\cancel{3}} \cdot \cancel{3}$ which means $3 {x}_{1} - 2 = 0$
$2 {x}_{1} = \frac{5}{\cancel{2}} \cdot \cancel{2}$ which means $2 {x}_{2} - 5 = 0$

Now, let's just factor these.

$\left(3 x - 2\right) \left(2 x - 5\right)$

Done.

May 15, 2015

$24 {x}^{2} - 76 x + 40 = 4 \cdot 6 {x}^{2} - 4 \cdot 19 x + 4 \cdot 10$

$= 4 \cdot \left(6 {x}^{2} - 19 x + 10\right)$

From the rational root theorem, if $\frac{p}{q}$ is a root of $6 {x}^{2} - 19 x + 10 = 0$ written in lowest terms, then $p$ must be a divisor of the constant term $10$ and $q$ must be a divisor of the coefficient ($6$) of the highest order term.

Notice that $6$ can be factored as $2 \cdot 3$ and $10$ can be factored as $2 \cdot 5$.

Let us try $\left(2 x - 5\right) \left(3 x - 2\right)$, it seems to work...

$\left(2 x - 5\right) \left(3 x - 2\right) = 2 x \cdot 3 x - 5 \cdot 3 x - 2 x \cdot 2 - 5 \cdot \left(- 2\right)$

$= 6 {x}^{2} - \left(15 + 4\right) x + 10$

$= 6 {x}^{2} - 19 x + 10$

So in summary, $24 {x}^{2} - 76 x + 40 = 4 \cdot \left(2 x - 5\right) \left(3 x - 1\right)$

May 15, 2015

$24 {x}^{2} - 76 x + 40 = 4 \left(6 {x}^{2} - 19 x + 10\right)$.

To factor $6 {x}^{2} - 19 x + 10$ multiply $6 \times 10 = 60$.

We need two numbers that multiply to give us $60$ and add to get us $- 19$ (the number in the middle).

A little thought should convince us that both numbers must be negative. We proceed in order:

$- 1 \times - 60$ do not add up to $- 19$

$- 2 \times - 30$ do not add up to $- 19$

$- 3 \times - 10$ do not add up to $- 19$

$- 4 \times - 15$ STOP! These do add up to $- 19$.

We'll split the $- 19 x$ into $- 4 x - 15 x$ (or $- 15 x - 4 x$ either will work)

$6 {x}^{2} - 19 x + 10 = 6 {x}^{2} - 4 x - 15 x + 10$.

Now factor by grouping:

$6 {x}^{2} - 19 x + 10 = \left(6 {x}^{2} - 4 x\right) + \left(- 15 x + 10\right)$.

$6 {x}^{2} - 19 x + 10 = 2 x \left(3 x - 3\right) + \left(- 5\right) \left(3 x - 2\right)$ .

$6 {x}^{2} - 19 x + 10 = \left(2 x - 5\right) \left(3 x - 2\right)$.

So we finish with:

$24 {x}^{2} - 76 x + 40 = 4 \left(6 {x}^{2} - 19 x + 10\right)$

$\textcolor{w h i t e}{\text{ssssssssssssssss}}$ $= 4 \left(2 x - 5\right) \left(3 x - 2\right)$