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MarjorieT

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Algebra factoring for quadratic equations

I am having the hardest time figuring this out, if x (two the second power)-6x-16=0, is the factor 3 or something else.

mikau

Member

Registered: 2005-08-22

Posts: 1,504

Re: Algebra factoring for quadratic equations

okay, first we usually write x to the second power as x^2. Basically, x^n is x to the nth power.

anyway, what you have then is you must factor

x^2 -6x -16 = 0,

the question you have to ask yourself is, what two numbers multiplied give you -16 and added equal -6?

the answer is, 2 and -8, because 2*(-8) = -16 and 2 + (-8) = -6, now that you have these two numbers, here's what you do
(x + firstNumber)*(x + secondNumber)

so the factored form is (x + 2 )(x + (-8)) = (x + 2)(x-8). Note you can stick the numbers in any order you want, so you could have put -8 first to get (x - 8)(x+2) which is the same answer and just as good.

Last edited by mikau (2007-11-06 09:34:10)

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MarjorieT

Member

Registered: 2007-11-06

Posts: 0

Re: Algebra factoring for quadratic equations

Thank  you for your help. How about this one 6x^2+42x=0

MarjorieT

Member

Registered: 2007-11-06

Posts: 0

Re: Algebra factoring for quadratic equations

if f(t)=2t^2-4t-1 what is  the solution for f2 and f-1

Daniel123

Member

Registered: 2007-05-23

Posts: 663

Re: Algebra factoring for quadratic equations

MarjorieT
Today 19:11:46 Thank  you for your help. How about this one 6x^2+42x=0

Take out the heighest common factor, which is 6x, so:
6x(x+7) = 0

MarjorieT
Today 19:19:48 if f(t)=2t^2-4t-1 what is  the solution for f2 and f-1

Just replace t with 2, and then with -1 so:

f(2) = 2(2)^2 - 4(2) - 1 = -1
f(-1) = 2(-1)^2 - 4(-1) - 1 = 5

Last edited by Daniel123 (2007-11-07 09:06:03)