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humorist-workshop
algebra 2 question
how would I solve the equation 2x^2+3x-5(4) The xs are all "x" not multiplication signs. Thanks
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I'm confused on why its -5(4)?
Anyways, I can show you how to solve an equation LIKE that
How about..
3x^2-9x+6=0
Okay, so you Multiply the 3x^2 by 6. Which is 18x^2
Then -9x is the other one.
find two numbers that add to -9x and multiply to 18x^2
They are -6x and -3x
So you plug those two numbers in for -9x
Then parenthesis around the first two numbers, and the last two. so it will look like this
(3x^2-6x) (-3x+6)
Then you take out what you can
so..
you can take 3x out of the first one and -3 out of the second one so it will look like
3x(x-2) and -3(x-2)
you have to make sure the two parenthesis have the same thing in it.
then you put (3x-3) (x-2)
and there you go.
You have it.
You could send me a message explaining the last equation and i could help you, i just didnt understand it
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Factoring.
You know that the first 2 numbers in each equation have to be x and 2x, and the last two have to be 4 and 5.
Now, you just need to discover where to add and subtract (you have to add for one and subtract for the other because the last term in the original equation is a negative). You also have to determine if the 4 goes with the x or 2x (and then put the 5 with the other)
So it could be:
(x+4)(2x-5)
(2x+4)(x-5)
(x-4)(2x+5)
(x+4)(2x-5)
From there, you just have to FOIL to see which one matches the original equation.
That's how to do it with shortcuts, anyway.
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