Is it possible to factor y= x^2 + 4x -21 ? If so, what are the factors?

Jun 18, 2018

$\left(x + 7\right) \left(x - 3\right)$

Explanation:

$\text{the factors of - 21 which sum to + 4 are + 7 and - 3}$

${x}^{2} + 4 x - 21 = \left(x + 7\right) \left(x - 3\right)$

Jun 18, 2018

Yes: $\left(x + 7\right) \left(x - 3\right)$

Explanation:

You can do it solving: ${x}^{2} + 4 x - 21 = 0$

${x}^{2} + 4 x - 21 = 0$

$a = 1 , b = 4 , c = - 21$

$x = \setminus \frac{- b \setminus \pm \setminus \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \setminus \frac{- \left(4\right) \setminus \pm \setminus \sqrt{{\left(4\right)}^{2} - 4 \left(1\right) \left(- 21\right)}}{2 \left(1\right)}$

$x = \setminus \frac{- 4 \setminus \pm \setminus \sqrt{16 + 84}}{2 \left(1\right)}$

$x = \setminus \frac{- 4 \setminus \pm \setminus \sqrt{100}}{2}$

$x = \setminus \frac{- 4 \setminus \pm 10}{2}$
then
${x}_{1} = \setminus \frac{- 4 + 10}{2} = \setminus \frac{6}{2} = 3$
${x}_{2} = \setminus \frac{- 4 - 10}{2} = \setminus \frac{- 14}{2} = - 7$
And

${x}^{2} + b x + c = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$
Or
${x}^{2} + 4 x - 21 = \left(x - 3\right) \left(x - \left(- 7\right)\right)$

${x}^{2} + 4 x - 21 = \left(x - 3\right) \left(x + 7\right)$
Q.E.D.