SOLUTION: Prove that x=2 is a real zero of y=-x^3+7x^2-55x+90

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Question 917247: Prove that x=2 is a real zero of y=-x^3+7x^2-55x+90
Found 2 solutions by MathLover1, jim_thompson5910:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
y=-x%5E3%2B7x%5E2-55x%2B90+ factor it first, factor out -1
y=-1%28x%5E3-7x%5E2%2B55x-90%29+............write -7x%5E2 as -2x%5E2-5x%5E2 and 55x as 45x%2B10x
y=-%28+x%5E3-5x%5E2%2B45x-2x%5E2%2B10x-90%29 ...group
y=-%28%28+x%5E3-5x%5E2%2B45x%29-%282x%5E2-10x%2B90%29%29+
y=-%28x%28+x%5E2-5x%2B45%29-2%28x%5E2-5x%2B45%29%29
y=-%28%28x-2%29%28x%5E2-5x%2B45%29%29 ...as you can see one of the factors is x-2 and if we set it equal to zero we get:
x-2=0 => x=2 which proves that x=2 is a real zero of y=-x%5E3%2B7x%5E2-55x%2B90

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
An alternative easier proof is to plug x = 2 into the equation and evaluate.


y=-x^3+7x^2-55x+90

y=-(2)^3+7(2)^2-55(2)+90

y=0


Since we get a result of y = 0, this means that x = 2 is indeed a root or zero of y=-x^3+7x^2-55x+90


Let me know if you need more help or if you need me to explain a step in more detail.
Feel free to email me at jim_thompson5910@hotmail.com
or you can visit my website here: http://www.freewebs.com/jimthompson5910/home.html

Thanks,

Jim