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 Friday 20th of September

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 Depdendent Variable

 Number of equations to solve: 23456789
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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Factoring Polynomials by Grouping

Example

Factor: 20w2 - 21y + 12w - 35wy

Solution

To avoid an error with the signs, write each subtraction as an addition of the opposite.

20w2 - 21y + 12w - 35wy

= 20w2 + (-21y) + 12w + (-35wy)

Step 1 Factor each term.

= 2 Â· 2 Â· 5 Â· w Â· w + (-3 Â· 7 Â· y) + 2 Â· 2 Â· 3 Â· w + (-5 Â· 7 Â· w Â· y)

Step 2 Group terms with common factors.

The first and third terms have common factors 2, 2, and w. Therefore, group those terms together.

The second and fourth terms have common factors 7 and y. Therefore, group those terms together.

= [2 Â· 2 Â· 5 Â· w Â· w + 2 Â· 2 Â· 3 Â· w] +  [(-3 Â· 7 Â· y) + (-5 Â· 7 Â· w Â· y)]

Step 3 In each group, factor out the GCF of the terms.

In the first group, factor out 2 Â· 2 Â· w.

= [2 Â· 2 Â· w(5 Â· w + 3)] + [(-3 Â· 7 Â· y) + (-5 Â· 7 Â· w Â· y)]

Simplify.

= [4w(5w + 3)] + [(-3 Â· 7 Â· y) + (-5 Â· 7 Â· w Â· y)]

In the second group, the common factor is 7y. However, both terms are negative. Factor out -7y, rather than +7y, so the terms of the binomial have the same sign as the binomial in the first group.

= [4w(5w + 3)] + [-7y(3 + 5w)]

In the second group, write 3 + 5w as 5w + 3 to match the binomial in the first group.

= [4w(5w + 3)] + [-7y(5w + 3)]

Step 4 Factor out the GCF of the polynomial.

Factor out the binomial (5w + 3).

= (5w + 3)(4w - 7y)

We have factored the given polynomial as follows:

20w2 - 21y + 12w - 35wy = (5w + 3)(4w - 7y)

You can multiply to check the factorization. We leave the check to you.

Within the brackets, the third and fourth terms have no common factor except 1. Therefore, group the third and fourth terms together and write each with factor 1.

Step 3 In each group, factor out the GCF of the terms.
 In the first group, factor out 2 Â· x. In the second group, factor out 1. = 3w[2 Â· x (x + 3 Â· y) + 1(x + 3 Â· y)]

Step 4 Factor out the GCF of the polynomial.

 Factor out (x + 3y). =3w[(x + 3y)(2x + 1)]

We can multiply to check the factorization.

 Is Is Is 3w[(x + 3y)(2x + 1)] 3w[2x2 + x + 6xy + 3y]  6wx2 + 3wx + 18wxy + 9wy = 6wx2 + 18wxy + 3wx + 9wy ? = 6wx2 + 18wxy + 3wx + 9wy ? = 6wx2 + 18wxy + 3wx + 9wy ? Yes