How do you factor the expressions x^2+3x-4?

Jun 22, 2018

${x}_{1} = - 4$
${x}_{2} = 1$,
You use the so called discriminant

Explanation:

Most of the quadratic equations have the form of: $a {x}^{2} + b x + c = 0$
In your case: a=1 , b=3, c=-4
Then we use the discriminant $D = {b}^{2} - 4 \cdot a \cdot c$
In this case:
$D = {3}^{2} - 4 \cdot 1 \cdot \left(- 4\right)$
We get the result: D=25
Now we need the roots, for which we use the equation:
${x}_{1} = \frac{- b - \sqrt{D}}{2 \cdot a}$
${x}_{2} = \frac{- b + \sqrt{D}}{2 \cdot a}$
We get:
${x}_{1} = - 4$
${x}_{2} = 1$
And that's all :)

Jun 22, 2018

$\left(x - 1\right) \left(x + 4\right)$

Explanation:

What tow numbers sum up to our middle term of $3$, and have a product of $- 4$ (the last term)?

After some trial and error, we arrive at $- 1$ and $4$. So we can factor our quadratic as

$\left(x - 1\right) \left(x + 4\right)$

Hope this helps!