# How do you factor: y= 3x^2+8x+5 ?

Feb 7, 2016

#### Answer:

(x + 1 )(3x + 5 )

#### Explanation:

To factor # y = ax^2 + bx + c

• find the factors of the product ac which also sum to b

• replace bx by these factors

for y $= 3 {x}^{2} + 8 x + 5$

a = 3 , b= 8 and c = 5

product ac = $3 \times 5 = 15$

require the factors of 15 which also sum to 8. These are 3 , 5

now rewrite as : $3 {x}^{2} + \left(3 x + 5 x\right) + 5$

factor in 'pairs' $\left[3 {x}^{2} + 3 x\right] \textcolor{b l a c k}{\text{ and }} \left[5 x + 5\right]$

$3 {x}^{2} + 3 x = 3 x \left(x + 1\right) \textcolor{b l a c k}{\text{ and }} 5 x + 5 = 5 \left(x + 1\right)$

there is now a common factor of (x + 1 )

hence : (x + 1 )(3x + 5 )

$\Rightarrow 3 {x}^{2} + 8 x + 5 = \left(x + 1\right) \left(3 x + 5\right)$