(Solved) Polynomial Factoring?

gaflyc90

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11 years ago

Anyone know how to factor polynomials because I am completely stumped? The tutor in our college algebra class yesterday..she wasn't the greatest, lets just put it that way. =) And my home work is due at 4 today, so can someone please help explain it to me. Like:

1.GCF of 8x^4, -24x^2
2.16p^6q^4, 32p^3q^3, -48pq^2
3.Factor: 8x^2-4x-20 (this one confuses me because there are no like terms. What do I do with it???

THANKS!

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dcgreen

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11 years ago

1. 8x^2 is GCF
8x^4 and -24x^4
Take the smaller exponent (2) and the GCF of 8 and 24 (8)

2. 16pq^2

3. 4(2x^2 - x - 5)
The trinomial does not factor.

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Answer accepted by topic starter

ddmmk

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Posts: 20

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11 years ago

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len5002

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11 years ago

1.
GCF of 8x^4, -24x^2 is 24 and x^4

2.
GCF of 16p^6q^4, 32p^3q^3, -48pq^2 is 96 and p^6q^12

3.
8x^2 - 4x - 20
= 4(4x^2 - x - 5)

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Leeta c.

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11 years ago

why don't you use the discriminant method?

So if you have a quadratic expression of the form ax^2 + bx + c and you want to factor it, you take the detla, ? = b^2 - 4ac, then you find the two solution : x1 = (-b - ??)/2a and x2 = (-b + ??)/2a
and then you an write the expression as a(x - x1)(x - x2)

For example, the last one is 8x^2-4x-20
? = 16 - 4*8*(-20) = 656

So x1 = ( -16 - ?656 ) / -8 = -2 - 1/2*?41 and x2 = -2 + 1/2*?41

Thus, 8x^2-4x-20 = 8 (x - (-2 - 1/2*?41)) (x - 2 + 1/2*?41)

** for the second one, 8x^4 - 24x^2 = x^2 (8x^2 - 24) = 8x^2 (x^2 - 3)

Now we factor x^2 - 3 using this identity: a^2 - b^2 = (a - b) (a + b)
In this case, a = x, and b = ?3

Thus 8x^2 - 24 = 8 x^2 ( x - ?3 ) (x + ?3 )

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