Factoring
6p^4-13p^3q^2-28p^3
I have come up with
(3p^2 + 4q)(2p^2 -7q^2q
Fix the typos and repost.
Find all numbers that must be excluded from the domain of the given rational expression.
1) x^2+4x/x^2-16
2) 6x^3y-6xy/x^3+5x^2+4x
3) 4-x^2/x^2+x-6
To factor the expression 6p^4-13p^3q^2-28p^3, we can start by looking for any common factors among the terms. In this case, the common factor is p^3.
First, let's factor out p^3:
p^3(6p-13q^2-28)
Now, we can focus on factoring the remaining expression inside the parentheses: 6p-13q^2-28.
There doesn't seem to be any common factors among the terms inside the parentheses, so we can apply the quadratic factoring method.
To factor the quadratic expression, we are looking for two binomials in the form (ap + bq)(cp + dq) that multiply to give us 6p-13q^2-28. The first term will involve the factors of 6p^2 (since 6p multiplied by p gives us 6p^2), and the last term will involve the factors of -28 (since -28 multiplied by 1 gives us -28).
So, let's start by listing the factor pairs of 6p^2 and -28:
Factor pairs of 6p^2:
(1p, 6p) or (2p, 3p) or (3p, 2p) or (6p, 1p)
Factor pairs of -28:
(1, -28) or (-1, 28) or (2, -14) or (-2, 14) or (4, -7) or (-4, 7) or (7, -4) or (-7, 4) or (14, -2) or (-14, 2) or (28, -1) or (-28, 1)
From these factor pairs, we need to find a combination that gives us a -13q^2 term when multiplied, as the middle term of our quadratic expression is -13q^2.
Looking at the factor pairs, the combination that satisfies this condition is (3p, -14) and (-2, 1). This is because (3p)(-2) gives us -6p, and (-14)(1) gives us -14, which adds up to -13p when combined.
So, we can rewrite the quadratic expression as:
p^3(3p - 14)(2q + 1)
Therefore, the factored form of the expression 6p^4-13p^3q^2-28p^3 is:
p^3(3p - 14)(2q + 1)