Solving Quadratic Equations by Factoring - Another Example
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Uploader Comments (patrickJMT)
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All Comments (10)
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this beats school by a mile ty so much
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I'm sorry but I need you ASAP, I have a problem just like 2nd step the beginning problem was -2/5(3X+1)(4X-3)=0 and i did your steps and I got -2/5X=-1/3 and -2/5X=+3/4 do I do anything to that "A" value or do I just leave it
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@jdjordan1985 thats what i need to know
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How do you factor 6x^2-36?
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I got it, thx.
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I love you
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that method is a bit long. just multiply out the brackets and incorporate the -81 giving y^2+12x-45 = 0
two numbers that add together giving 12 and multiply together giving -45: 15 & -3
hence factorizes to (y-3)(y+15)
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Um, I don't wanna waste your time, but I'm a little stuck halfway through the procedure. Here's my problem 23p=5p^2 + 24 Here's the first step I took.
0=5p^2+24-23p And that's as far as I've gotten. I don't know where to go from here.
Vospader21 11 months ago
@Vospader21 i would use the quadratic formula to solve this. otherwise it is just tedious trial/error factoring if it does factor with 'nice' numbers
patrickJMT 11 months ago