# Can y=2x^2 – 3x – 5 be factored? If so what are the factors ?

Apr 9, 2018

$x = \frac{5}{2} , - 1$

#### Explanation:

$2 {x}^{2} - 3 x - 5$

=$\left(2 x - 5\right) \left(x + 1\right)$

$x = \frac{5}{2} , - 1$

Apr 9, 2018

$\left(x + 1\right) \left(2 x - 5\right)$

#### Explanation:

$\text{for a quadratic in "color(blue)"standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{consider the factors of the product ac which sum to b}$

$\text{here "a=2,b=-3" and } c = - 5$

$\Rightarrow a c = 2 \times - 5 = - 10 \text{ and - 5, + 2 sum to - 3}$

$\textcolor{b l u e}{\text{split the middle term using these factors}}$

$\Rightarrow y = 2 {x}^{2} + 2 x - 5 x - 5 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

$\textcolor{w h i t e}{y} = \textcolor{red}{2 x} \left(x + 1\right) \textcolor{red}{- 5} \left(x + 1\right)$

$\text{take out a "color(blue)"common factor } \left(x + 1\right)$

$= \left(x + 1\right) \left(\textcolor{red}{2 x - 5}\right)$

$\Rightarrow 2 {x}^{2} - 3 x - 5 = \left(x + 1\right) \left(2 x - 5\right)$

Apr 9, 2018

$y = \left(2 x - 5\right) \left(x + 1\right)$

#### Explanation:

Can the trinomial be factored? $y = a {x}^{2} + b x + c$

Test it using $\text{ "b^2 -4ac" }$ first.
If the answer is a perfect square then you can find factors. This means you do not waste time by trying to find factors if there are none!

${\left(- 3\right)}^{2} - 4 \left(2\right) \left(- 5\right) = 9 + 40 = 49 \text{ } \leftarrow = {7}^{2}$

Find factors of $2 \mathmr{and} 5$ whose products differ by $3$

Both $2 \mathmr{and} 5$ are prime numbers so there are only two combinations to try:

$\left(2 \times 5\right) - \left(1 \times 1\right) = 10 - 1 = 9$

$\left(1 \times 5\right) - \left(2 \times 1\right) = 5 - 2 = 3$

$y = \left(2 x - 5\right) \left(x + 1\right)$