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

1 Answer
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 # = 3x^2 + 8x + 5#

a = 3 , b= 8 and c = 5

product ac = # 3 xx 5 = 15#

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

now rewrite as : # 3x^2 +( 3x + 5x) + 5#

factor in 'pairs' # [ 3x^2 + 3x ] color(black)(" and ") [5x + 5 ]#

# 3x^2 + 3x = 3x(x+ 1 ) color(black)(" and ") 5x + 5 = 5(x + 1 )#

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

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

# rArr 3x^2 + 8x + 5 = (x + 1 )(3x + 5 ) #