232=(3x+5)(x)

Rewrite the equation as (3x+5)(x)=232.

(3x+5)(x)=232

Apply the distributive property.

3x⋅x+5x=232

Multiply x by x by adding the exponents.

Move x.

3(x⋅x)+5x=232

Multiply x by x.

3×2+5x=232

3×2+5x=232

3×2+5x=232

Move 232 to the left side of the equation by subtracting it from both sides.

3×2+5x-232=0

For a polynomial of the form ax2+bx+c, rewrite the middle term as a sum of two terms whose product is a⋅c=3⋅-232=-696 and whose sum is b=5.

Factor 5 out of 5x.

3×2+5(x)-232=0

Rewrite 5 as -24 plus 29

3×2+(-24+29)x-232=0

Apply the distributive property.

3×2-24x+29x-232=0

3×2-24x+29x-232=0

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

(3×2-24x)+29x-232=0

Factor out the greatest common factor (GCF) from each group.

3x(x-8)+29(x-8)=0

3x(x-8)+29(x-8)=0

Factor the polynomial by factoring out the greatest common factor, x-8.

(x-8)(3x+29)=0

(x-8)(3x+29)=0

If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.

x-8=0

3x+29=0

Set the first factor equal to 0.

x-8=0

Add 8 to both sides of the equation.

x=8

x=8

Set the next factor equal to 0.

3x+29=0

Subtract 29 from both sides of the equation.

3x=-29

Divide each term by 3 and simplify.

Divide each term in 3x=-29 by 3.

3×3=-293

Cancel the common factor of 3.

Cancel the common factor.

3×3=-293

Divide x by 1.

x=-293

x=-293

Move the negative in front of the fraction.

x=-293

x=-293

x=-293

The final solution is all the values that make (x-8)(3x+29)=0 true.

x=8,-293

Solve Using the Square Root Property 232=(3x+5)(x)