Solving Quadratic Equations by Factoring - Another Example

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Uploaded by on May 25, 2010

Solving Quadratic Equations by Factoring - Another Example

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Uploader Comments (patrickJMT)

  • 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 i would use the quadratic formula to solve this. otherwise it is just tedious trial/error factoring if it does factor with 'nice' numbers

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  • @Vospader21 5p^2-23p+24

    5p^2-15p-8p+24

    5p(p-3)-8(p-3)

    (5p-8)(p-3)

  • this beats school by a mile ty so much

  • 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

  • @jdjordan1985 thats what i need to know

  • How do you factor 6x^2-36?

  • I got it, thx.

  • I love you

  • 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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