Expert: Sherman D. - 5/17/2004

Question
how do u solve 2x^2+20+8
2a^2-14a+24
3y^3+15y^2+12y


Answer
if by this you mean

2x^2 + 20 + 8
you really do mean that, then
2x^2 + 20 + 8
2x^2 + 28
2(x^2 + 14)
If you know what complex values are, then
2(x + isqrt(14))*(x - isqrt(14))
whereas "i" = sqrt(-1)

but if you meant

2x^2 + 20x + 8, then
2(x^2 + 10x + 4)
Can't factor any further

However if you meant

2x^2 + 2x + 8, then
2(x^2 + x + 4)
can't factor any further.

If i have any of that typed wrong, just let me know and i will fix it.

-----------------------------------------------------------

2a^2 - 14a + 24
2(a^2 - 7a + 12)
2(a - 4)(a - 3)

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3y^3 + 15y^2 + 12y
3y(y^2 + 5y + 4)
3y(y + 1)(y + 4)

or you can go to www.quickmath.com for help with factoring.