factor by grouping

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Factor By Grouping:

Factor By Grouping

Problem 1 : Factor 5(x + 8) + 3x(x + 8):

Problem 1 : Factor 5(x + 8) + 3x(x + 8 ) Factor 5(x + 8) + 3x(x + 8 ) In order to factor this expression, we need to find the common factor in each term.

Problem 1 : Factor 5(x + 8) + 3x(x + 8):

Problem 1 : Factor 5(x + 8) + 3x(x + 8) Factor 5 (x + 8) + 3x (x + 8) The binomial (x+8) is our GCF, so we need to factor it out of each term.

Problem 1: Answer:

Problem 1: Answer 5 (x + 8) + 3x (x + 8 ) (x + 8 )(5+3x) Once we factor (x+8) out of each term, the leftovers are grouped together as the remaining factor.

You Try…:

You Try… Factor: 2x (x – 5) + 3 (x -5) x 2 (y – 3) – 8 (y – 3)

Answers…:

Answers… Factor: 2x (x – 5) + 3 (x -5 ) = (2x + 3)(x – 5) x 2 (y – 3) – 8 (y – 3 ) = (x 2 – 8)(y – 3)

Remember, invisible coefficient 1:

Remember, invisible coefficient 1 x(13 + x) – (x + 13 ) In this problem, we need to remember that there is an understood 1 in front of the second group of parentheses. When we pull out the common factor (x+13), our leftovers are x and -1. Therefore, our answer is ( x - 1)(13 + x ).

Problem 2: Factor y3 – 4y2 + 8y - 32:

Problem 2: Factor y 3 – 4y 2 + 8y - 32 Factor y 3 – 4y 2 + 8y – 32 First we need to group the terms  into pairs.     

Problem 2: Factor y3 – 4y2 + 8y - 32:

Problem 2: Factor y 3 – 4y 2 + 8y - 32 ( y 3 – 4y 2 )+ (8y – 32) Secondly, we need to factor out a GCF from each pair.

Problem 2: Factor y3 – 4y2 + 8y - 32:

Problem 2: Factor y 3 – 4y 2 + 8y - 32 y 2 (y – 4) + 8(y – 4) Now that we have the GCFs, we finish by f actoring out the common binomial (y-4).

Problem 2: Answer:

Problem 2: Answer ( y 2 + 8)(y – 4)

Your Try…:

Your Try… Factor: 5w + 10 + 3xw + 6x 2wx + 10w + 7x + 35 a 3 + 6a – 5a 2 – 30

Answers….:

Answers…. Factor: 5w + 10 + 3xw + 6x = (5 + 3x)(w + 2) 2wx + 10w + 7x + 35 = ( 2w + 7)(x + 5) a 3 + 6a – 5a 2 – 30 = ( a – 5)(a + 6)