# How do you find the zeros by rewriting the function y=x^2+4x+3 in intercept form?

Aug 23, 2017

$x = - 3 , x = - 1$

#### Explanation:

$\text{factorise the quadratic } {x}^{2} + 4 x + 3$

$\text{the product of the factors of +3 which sum to + 4}$
$\text{are + 1 and + 3}$

$\Rightarrow y = \left(x + 1\right) \left(x + 3\right) \leftarrow \textcolor{red}{\text{ in intercept form}}$

$\text{to find the zeros equate y to zero}$

$\Rightarrow \left(x + 1\right) \left(x + 3\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x + 1 = 0 \Rightarrow x = - 1$

$x + 3 = 0 \Rightarrow x = - 3$
graph{x^2+4x+3 [-10, 10, -5, 5]}