# Is it possible to factor y=2x^2+7x+3 ? If so, what are the factors?

Dec 3, 2015

Yes, it is possible.

$y = \left(2 {x}^{2} + 7 x + 3\right) = \left(x + \frac{1}{2}\right) \left(x + 3\right)$

#### Explanation:

To find a complete factorisation of $2 {x}^{2} + 7 x + 3$, you need to find all solutions of the equation

$2 {x}^{2} + 7 x + 3 = 0$.

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This can be done e.g. with "completion of the circle".
Let me show you how it's done!

... compute $- 3$ on both sides of the equation...

$\iff 2 {x}^{2} + 7 x = - 3$

... divide by $2$ on both sides of the equation ...

$\iff {x}^{2} + \frac{7}{2} x = - \frac{3}{2}$

Now, the goal is to create a quadratic expression like ${x}^{2} + 2 a x + {a}^{2}$ on the left side of the equation.

We already have ${x}^{2}$ and $\frac{7}{2} x = 2 a x$ which means that $a = \frac{7}{4}$. So, to complete the circle, we need to add ${\left(\frac{7}{4}\right)}^{2}$ to the expression on the left side.
However, as we don't want to destroy the equality, it is also needed to add the same expression on the right side, too.

$\iff {x}^{2} + \frac{7}{2} x + {\left(\frac{7}{4}\right)}^{2} = - \frac{3}{2} + {\left(\frac{7}{4}\right)}^{2}$

As next, due to the rule ${x}^{2} + 2 a x + {a}^{2} = {\left(x + a\right)}^{2}$, you can transform the left side.

$\iff {\left(x + \frac{7}{4}\right)}^{2} = - \frac{3}{2} + {\left(\frac{7}{4}\right)}^{2}$

$\iff {\left(x + \frac{7}{4}\right)}^{2} = \frac{25}{16}$

Now, you can draw the root on both sides.
Be careful though: you will have two solutions since both ${\left(\frac{5}{4}\right)}^{2} = \frac{25}{16}$ and ${\left(- \frac{5}{4}\right)}^{2} = \frac{25}{16}$ hold.

$x + \frac{7}{4} = \frac{5}{4} \textcolor{w h i t e}{\times} \text{ or } \textcolor{w h i t e}{\times} x + \frac{7}{4} = - \frac{5}{4}$

$x = - \frac{1}{2} \textcolor{w h i t e}{\times \xi i i} \text{ or } \textcolor{w h i t e}{\times} x = - 3$

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With your solutions $x = - \frac{1}{2}$ and $x = - 3$ you have:

$\textcolor{w h i t e}{\times x} 2 {x}^{2} + 7 x + 3 = 0$

$\iff \left(x - \left(- \frac{1}{2}\right)\right) \left(x - \left(- 3\right)\right) = 0$

$\iff \left(x + \frac{1}{2}\right) \left(x + 3\right) = 0$

Thus,

$2 {x}^{2} + 7 x + 3 = \left(x + \frac{1}{2}\right) \left(x + 3\right)$.