Chapter 1.3 Algebraic expressions

# Find and factor out the gcf of the variable parts of

• Notes
• 26

This preview shows page 21 - 25 out of 26 pages.

Find and factor out the GCF of the variable parts of the monomials. Factor the remaining polynomial. Example 22: Factor 60 x 2 y 3 z 2 + 120 x 2 y 3 z + 60 x 2 y 3 Solution: The expression to be factored is a sum of 3 terms. Preliminary work: Find the two GCF’s The GCF of 60 = 2 2 · 3 · 5 , 120 = 2 3 · 3 · 5 and 60 = 2 2 · 3 · 5 is 2 2 · 3 · 5 = 60 The GCF of the variable parts of the 3 terms is clearly x 2 y 3 (product of lowest powers that appear in all terms). Now put everything together: Original problem: Factor 60 x 2 y 3 z 2 + 120 x 2 y 3 z + 60 x 2 y 3 1. GCF of coefficients is 60; factor it out: = 60( x 2 y 3 z 2 + 2 x 2 y 3 z + x 2 y 3 ) 2. GCF of variable parts is x 2 y 3 ; factor it out: = 60 x 2 y 3 ( z 2 + 2 z + 1) 3. Factor the remaining polynomial z 2 + 2 z + 1 = 60 x 2 y 3 ( z + 1) 2 Stanley Ocken M19500 Precalculus Chapter 1.3: Algebraic expressions

Subscribe to view the full document.

Welcome Algebraic Expressions Simplified sums Greatest common factor Factoring Exercises Quiz Review Nothing changes when you factor expressions with fractional exponents: just be careful to figure out correctly which (fractional) exponent is the lowest! Example 23: Factor 12 x 1 / 2 - 36 x - 1 / 2 + 24 x - 3 / 2 Solution: Here is the original problem 12 x 1 / 2 - 36 x - 1 / 2 + 24 x - 3 / 2 Factor out GCF of 12 , 36 , 24 : = 12( x 1 / 2 - 3 x - 1 / 2 + 2 x - 3 / 2 ) Factor out lowest power of x : = 12 x - 3 2 ( x 1 / 2 - ( - 3 / 2) - 3 x - 1 / 2 - ( - 3 / 2) + 2 x - 3 / 2 - ( - 3 / 2) ) = 12 x - 3 / 2 ( x 2 - 3 x 1 + 2 x 0) Factor the remaining part: = 12 x - 3 / 2 ( x 2 - 3 x + 2) Here is the answer: = 12 x - 3 / 2 ( x - 1)( x - 2) Example 15 in SRW 1.3 is an artificial method of factoring, and so we omit it. Stanley Ocken M19500 Precalculus Chapter 1.3: Algebraic expressions
Welcome Algebraic Expressions Simplified sums Greatest common factor Factoring Exercises Quiz Review Example 24: Factor 40( x - 1) 3 (2 x + 1) 4 - 24( x - 1) 4 (2 x + 1) 3 (*) Step 1. The GCF of the coefficients 40 = 2 3 · 5 and 24 = 2 3 · 3 is the product of lowest prime powers 2 3 · 3 0 · 5 0 = 8 . Step 2: The GCF of the variable parts ( x - 1) 3 (2 x + 1) 4 and ( x - 1) 4 (2 x + 1) 3 is the product of lowest powers ( x - 1) 3 (2 x + 1) 3 Step 3: First factor out 8 from (*); then factor out ( x - 1) 3 (2 x + 1) 3 Solution: Expression to factor is 40( x - 1) 3 (2 x + 1) 4 - 24( x - 1) 4 (2 x + 1) 3 Factor out GCF of coefficients: = 8 (5( x - 1) 3 (2 x + 1) 4 - 3( x - 1) 4 (2 x + 1) 3 ) Factor out GCF of variable parts: = 8( x - 1) 3 (2 x + 1) 3 (5(2 x + 1)) - (3( x - 1)) Expand the cofactor = 8( x - 1) 3 (2 x + 1) 3 (10 x + 5 - 3 x + 3) Simplify the cofactor = 8( x - 1) 3 (2 x + 1) 3 (7 x + 8) Stanley Ocken M19500 Precalculus Chapter 1.3: Algebraic expressions

Subscribe to view the full document.

Welcome Algebraic Expressions Simplified sums Greatest common factor Factoring Exercises Quiz Review Exercises for Chaper 1.3 Click on Wolfram Calculator to find an answer checker. Click on Wolfram Algebra Examples to see how to check various types of algebra problems. 1. Do the WebAssign exercises for Chapter 1.3. Rewrite each product as a simplified sum: 2. ( x + 7)( x - 7) 3. 2(2 x + 1)( x - 1) 2 4. ( x + 2)( x 2 + x + 1) 5. Factor each polynomial completely: a) x 2 - 17 x + 60 b) x 2 - 10 x - 75 c) 4 x 4 - 24 x 3 + 32 x 2 d) y 4 ( y + 2) 4 - y 3 ( y + 2) 5 e) 5( x 2 + 4) 4 (2 x )( x - 2) 4 + ( x 2 + 4) 5 (4)( x - 2) 3 f) ( x 2 + 3) - 1 / 3 - 2 3 x 2 ( x 2 + 3) - 4 / 3 g) 100 x 2 yz + 125 xyz 2 + 200 xy 2 z h) 40 x 3 ( x - 1) 7 (2 x + 1) 4 - 24( x - 1) 4 x 3 (2 x + 1) 8 .

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern