53 7.2 Factoring by Grouping
First thing to do when factoring is to factor out the GCF. This GCF is often a monomial, like in the problem \(5xy + 10xz\) where the GCF is the monomial \(5x\), so you would have \(5x(y + 2z)\). However, a GCF does not have to be a monomial; it could be a binomial. Consider the following two examples.
Example 7.2.1
Find and factor out the GCF for \(3ax - 7bx\).
By observation, one can see that both have \(x\) in common.
This means that \(3ax - 7bx = x(3a - 7b)\).
Example 7.2.2
Find and factor out the GCF for \(3a(2a + 5b) - 7b(2a + 5b)\).
Both have \((2a + 5b)\) as a common factor.
This means that if you factor out \((2a + 5b)\), you are left with \(3a - 7b\).
The factored polynomial is written as \((2a + 5b)(3a - 7b)\).
In the same way as factoring out a GCF from a binomial, there is a process known as grouping to factor out common binomials from a polynomial containing four terms.
Find and factor out the GCF for \(10ab + 15b^2 + 4a + 6b\).
To do this, first split the polynomial into two binomials.
\(10ab + 15b^2 + 4a + 6b\) becomes \(10ab + 15b^2\) and \(4a + 6b\).
Now find the common factor from each binomial.
\(10ab + 15b^2\) has a common factor of \(5b\) and becomes \(5b(2a + 3b)\).
\(4a + 6b\) has a common factor of 2 and becomes \(2(2a + 3b)\).
This means that \(10ab + 15b^2 + 4a + 6b = 5b(2a + 3b) + 2(2a + 3b)\).
\(5b(2a + 3b) + 2(2a + 3b)\) can be factored as \((2a + 3b)(5b + 2)\).
Questions
Factor the following polynomials.
- \(40r^3-8r^2-25r+5\)
- \(35x^3-10x^2-56x+16\)
- \(3n^3-2n^2-9n+6\)
- \(14v^3+10v^2-7v-5\)
- \(15b^3+21b^2-35b-49\)
- \(6x^3-48x^2+5x-40\)
- \(35x^3-28x^2-20x+16\)
- \(7n^3+21n^2-5n-15\)
- \(7xy-49x+5y-35\)
- \(42r^3-49r^2+18r-21\)
- \(16xy-56x+2y-7\)
- \(3mn-8m+15n-40\)
- \(2xy-8x^2+7y^3-28y^2x\)
- \(5mn+2m-25n-10\)
- \(40xy+35x-8y^2-7y\)
- \(8xy+56x-y-7\)
- \(10xy+30+25x+12y\)
- \(24xy+25y^2-20x-30y^3\)
- \(3uv+14u-6u^2-7v\)
- \(56ab+14-49a-16b\)
