How Do You Solve the Quadratic Equation (x+3)(x-2)^2=72?

  • Thread starter ihopeican
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    Algebra Hi
In summary, To solve the given quadratic equation, multiply each term of one binomial by each term of the other, and then combine like terms. This will result in x^2 + x - 6. Then add 72 and set the equation equal to 0. Finally, use the quadratic formula to find the solution and check for 0.
  • #1
ihopeican
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hi there,

i just forgot how to solve quadratics and was wondering if anyone could help me on a question it is quite easy.

(x+3)(x-2)^2= 72
(x-2)^2
=x^2-4x+4
(x+3)(x^2-4x+4)
x^3-4x^2+4x+3x^2-12x+12

=x^3-x^2-8x+12= 72

im not quite sure what i should do next if i haven't made any mistakes so far.

thankyou,
 
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  • #2
I would start by noting 72=9*8=3*3*8.

This should get you one solution.

Then when you get to

x^3-x^2-8x+12= 72

Subtract to get

x3-x2-8x-60=0

and factor out x-a, where a is the solution you found earlier.
 
  • #3
Guess factors of 60 to plug in using synthetic division/substitution
 
  • #4
multiply each term of one binomial by each term of of the other

(x + 3)(x-2)^2=72

multiply each term by 3
3 * -2 = -6
3 * x = 3x

combine like terms
x^2 + x3 - x2 -6

is

x^2 + x -6

now

x^2 + x -6 + 72

do the quadratic eq and check for 0
 

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