# 20 x 3 24 x 2 15 x 18 4 x 2 5 x 6 35 x 6 4 x 2 3 5 x

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20 x 3 + 24 x 2 + 15 x + 18 = 4 x 2 (5 x + 6) + 3(5 x + 6) = (4 x 2 + 3) (5 x + 6). Solution using table approach: Put the four terms in the four inside cells of a distributive property multiplication table. Then try to work out a set of expressions that could go on the outside of the table. If you can do this successfully, you’ve found the factorization. ? ? ? 20 x 3 24 x 2 ? 15 x 18 5 x 6 4 x 2 20 x 3 24 x 2 3 15 x 18 20 x 3 + 24 x 2 + 15 x + 18 = (4 x 2 + 3) (5 x + 6). You try it 1. Factor 7 x 3 – 14 x 2 – 3 x + 6. 2. Factor 6 x 3 + 15 x 2 – 2 x – 5.

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Name Honors Algebra 2 February 24, 2012 “Advanced factoring methods” page 2 B. Factoring by variable substitution Some polynomials can be factored by substituting a variable for a power of x , creating a simpler factoring problem. Eventually the final answer needs to be written in terms of the original variable. Example: Factor x 4 – 10 x 2 + 9. Solution: Use the substitution u = x 2 . x 4 – 10 x 2 + 9 = ( x 2 ) 2 – 10 x 2 + 9 = u 2 – 10 u + 9 = ( u – 1)( u – 9) = ( x 2 – 1)( x 2 – 9) = ( x +1)( x –1)( x +3)( x –3). This substitution method is often useful in equation solving problems.

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