Well we are doing factoring and I don't quite understand this problem:
(x+1)(x-3)=117
I simplified and got:
x^2-3x+x-3=117
x^2-2x-3=117
x^2-2x-120=0
I split the middle term and got:
x^2-12x+10x-120=0
What should I do next? I have no idea how to finish factoring this...
thanks,
meloney
now you want to factor by "grouping"
x^2-12x+10x-120=0
x(x-12) + 10(x-12) = 0
((x-12)(x+10) = 0
so x= 12 or x = -10
wow I never would have thought of grouping them. Thanks!!!
To factor the quadratic equation x^2-2x-120=0, you can try using the method of factoring by grouping or the quadratic formula. Let's first explain the process of factoring by grouping:
1. Begin by grouping the terms as follows:
(x^2-12x) + (10x-120) = 0
2. Factor out the greatest common factor from each group:
x(x-12) + 10(x-12) = 0
3. Notice that you now have a common factor of (x-12) in each term. Factor it out:
(x-12)(x+10) = 0
4. Apply the zero product property, which states that if the product of two factors is equal to zero, then at least one of the factors must be equal to zero:
x-12 = 0 or x+10 = 0
5. Solve for x in each equation:
x = 12 or x = -10
So, the solutions to the quadratic equation x^2-2x-120=0 are x = 12 and x = -10.
To check if these solutions are correct, substitute them back into the original equation:
For x = 12:
(x+1)(x-3) = 117
(12+1)(12-3) = 117
13*9 = 117
117 = 117 (true)
For x = -10:
(-10+1)(-10-3) = 117
(-9)(-13) = 117
117 = 117 (true)
Therefore, x = 12 and x = -10 are the correct solutions to the equation.