How do you solve #3x^2+5x+2=0# by factoring?

Question

Hazel Anglin · Accepted Answer

Assuming #3x^2+5x+2# can be factored into
#(ax+b)(cx+d)# with integer values for #a,b,c,d#
and noting that all of #a,b,c,d# will be non-negative (from the form of the original quadratic)
we only have the possibilities
#{a,c} = {1,3}# and
#{b,d} = {1,2}# There are only 2 combinations to attempt:
#(1x+1)(3x+2)#
and
#(1x+2)(3x+1)# revealing that #3x^2+5x+2 =0# can be factored as
#(x+1)(3x+2)=0# So either
#x+1 = 0#  which implies #x = -1#
or
#3x+2 = 0# which implies #x = -2/3#

Avery Alexander · Answer

undefined To solve the quadratic equation $3x^2 + 5x + 2 = 0$ by factoring, follow these steps:

1. Write down the equation: $3x^2 + 5x + 2 = 0$.
2. Factor the quadratic expression $3x^2 + 5x + 2$ into two binomial factors.
3. Set each factor equal to zero and solve for $x$.
4. The solutions obtained will be the roots of the quadratic equation.

Factoring the quadratic expression $3x^2 + 5x + 2$:

$3x^2 + 5x + 2 = (3x + 2)(x + 1)$.

Setting each factor equal to zero:

$3x + 2 = 0$ and $x + 1 = 0$.

Solving for $x$:

For $3x + 2 = 0$:  
$3x = -2$  
$x = -\frac{2}{3}$.

For $x + 1 = 0$:  
$x = -1$.

Therefore, the solutions to the quadratic equation $3x^2 + 5x + 2 = 0$ are $x = -\frac{2}{3}$ and $x = -1$.