University:
Glendale Community College
Course:
MAT141 | College Mathematics
Academic year:

2020
Views:

420

Pages:

15
Author:

Lorenzo Jarvis

Section 5.6: Factoring Sums and Differences of Cubes Chapter 5: Factoring Difference of Cubes After any GCF is factored out, an expression is considered a difference of cubes provided: • • • • Only has 2 terms Minus sign between the terms Variables have exponents that are multiples of 3, such as 𝑥 3 , 𝑦 6 , 𝑧 9 , 𝑤 12 … Numbers are perfect cubes, such as 1, 8, 27, 64, 125, 216, 343, 512, … If all 4 above are true after any GCF is factored out, then we factor as follows: Step 1: Factor out GCF, if applicable Step 2: Use the rule: 𝑎3 − 𝑏 3 = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) Step 3: Determine “a”. • • Cube root any numbers in the first term. (if applicable) Divide the exponent of the first term’s variable by 3. (if applicable) Step 4: Determine “b” • • Cube root any numbers in the first term. (if applicable) Divide the exponent of the first term’s variable by 3. (if applicable Step 5: Place the numbers in the formula and simplify to get the answer. Step 6: Check Section 5.6: Factoring Sums and Differences of Cubes Chapter 5: Factoring Sum of Cubes After any GCF is factored out, an expression is considered a sum of cubes provided: • • • • Only has 2 terms Plus sign between the terms Variables have exponents that are multiples of 3, such as 𝑥 3 , 𝑦 6 , 𝑧 9 , 𝑤 12 … Numbers are perfect cubes, such as 1, 8, 27, 64, 125, 216, 343, 512, … If all 4 above are true after any GCF is factored out, then we factor as follows: Step 1: Factor out GCF, if applicable Step 2: Create template: 𝑎3 + 𝑏 3 = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 ) Step 3: Determine “a”. • • Cube root any numbers in the first term. (if applicable) Divide the exponent of the first term’s variable by 3. (if applicable) Step 4: Determine “b” • • Cube root any numbers in the first term. (if applicable) Divide the exponent of the first term’s variable by 3. (if applicable Step 5: Place the numbers in the formula and simplify to get the answer. Step 6: Check Section 5.6: Factoring Sums and Differences of Cubes For Example: factor 27𝑥 3 − 64 Chapter 5: Factoring Solution: There is no GCF. Check to see if it is a difference of cubes: Step 1: Factor out GCF, if applicable Step 2: Create Template: 𝑎3 − 𝑏 3 • • • = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) Step 3: Determine “a”. • • • Cube root any numbers in the first term. (if applicable) Divide the exponent of the first term’s variable by 3. (if applicable) Step 4: Determine “b” • • Cube root any numbers in the first term. (if applicable) Divide the exponent of the first term’s variable by 3. (if applicable Step 5: Place the numbers in the formula and simplify to get the answer. Step 6: Check Only has 2 terms ✓ Minus sign between the terms ✓ Variables have exponents that are multiples of 3, 𝑥 3 , 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 3 𝑖𝑠 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3 ✓ Numbers are perfect cubes, 27 = 33 , 64 = 43 ✓ Problem is a difference of cubes. Step 1: There is no GCF, skip to step 2. Step 2: Create Template: (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) 3 Step 3: Determine “a”. 𝑎 = √27𝑥 3⁄3 = 3𝑥 𝑎 = 3𝑥 3 Step 3: Determine “b”. 𝑏 = √64 = 4 b=4 Step 5: Place the numbers in the formula and simplify to get the answer. 27𝑥 3 − 64 = (3𝑥 − 4)((3𝑥)2 + (3𝑥)(4) + (4)2 ) = Answer: (3𝑥 − 4)(9𝑥 2 + 12𝑥 + 16) Step 6: Check 9𝑥 2 + 12𝑥 + 16 × 3𝑥 − 4 −36𝑥 2 − 48𝑥 − 64 27𝑥 3 + 36𝑥 2 + 48𝑥___ 27𝑥 3 − 64 ✓ Section 5.6: Factoring Sums and Differences of Cubes For Example: Factor 125𝑥 3 + 𝑦 6 Chapter 5: Factoring Solution: There is no GCF. Check to see if it is a sum of cubes: Step 1: Factor out GCF, if applicable Step 2: Create Template: 𝑎3 + 𝑏 3 • • • Only has 2 terms ✓ Plus sign between the terms ✓ Variables have exponents that are multiples of 3, 𝑥 3 , 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 3 𝑖𝑠 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 3 𝑦 6 exponent of 6 is a multiple of 3. ✓ • Numbers are perfect cubes, 125= 53 , ✓ = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 ) Step 3: Determine “a”. • • Cube root any numbers in the first term. (if applicable) Divide the exponent of the first term’s variable by 3. (if applicable) Problem is a sum of cubes. Step 1: There is no GCF, skip to step 2. Step 2: Create Template: (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 ) 3 Step 4: Determine “b” • • Cube root any numbers in the first term. (if applicable) Divide the exponent of the first term’s variable by 3. (if applicable Step 5: Place the numbers in the formula and simplify to get the answer. Step 6: Check Step 3: Determine “a”. 𝑎 = √125𝑥 3⁄3 = 5𝑥 𝑎 = 3𝑥 Step 3: Determine “b”. 𝑏 = 𝑦 6⁄3 = 𝑦 2 b = 𝑦2 Step 5: Place the numbers in the formula and simplify to get the answer. 125𝑥 3 + 𝑦 6 = (5𝑥 + 𝑦 2 )((5𝑥)2 − (5𝑥)(𝑦 2 ) + (𝑦 2 )2 ) = Answer: (5𝑥 + 𝑦 2 )(25𝑥 2 − 5𝑥𝑦 2 + 𝑦 4 ) Step 6: Check 25𝑥 2 − 5𝑥𝑦 2 + 𝑦 4 × 5𝑥 + 𝑦 2 25𝑥 2 𝑦 2 − 5𝑥𝑦 4 + 𝑦 6 125𝑥 3 − 25𝑥 2 𝑦 2 + 5𝑥𝑦 4 ___ 125𝑥 3 + 𝑦 6 ✓ Section 5.6: Factoring Sums and Differences of Cubes Factoring Chapter 5: #1-42: Completely factor the binomials, remember to factor out the GCF first when applicable (if a problem is prime say so). 1) 𝑥 3 + 8 2) 𝑦 3 + 64 3) 𝑥 3 – 8 4) 𝑦 3 – 64 5) 𝑏 3 + 27 6) 𝑥 3 + 125 7) 𝑏 3 – 27 8) 𝑥 3 – 125 9) 𝑥 3 + 64 10) 𝑦 3 + 27 11) 𝑥 3 − 64 12) 𝑦 3 − 27 13) 8𝑥 3 – 27 14) 125𝑥 3 – 64 15) 8𝑥 3 + 27 16) 125𝑥 3 + 64 17) 27𝑥 3 – 125 18) 64𝑥 3 – 27 19) 64𝑥 3 – 𝑦 3 20) 27𝑥 3 – 𝑦 3 21) x9 – y3 22) 𝑦 6 – 𝑥 3 23) 2𝑥 6 – 54 24) 3𝑥 6 – 81 25) 125x9 – y6 26) 8𝑦 9 – 125𝑥 6 27) 16𝑥 3 – 54 28) 3𝑥 3 – 81 29) 3𝑥 3 + 24 30) 2𝑥 3 + 128 31) 𝑥 4 – 8𝑥 32) y4 – 64y 33) 6𝑥 4 – 48𝑥 34) 3𝑦 4 – 192𝑦 35) 8𝑥 5 + 125𝑥 2 36) 3𝑥 5 – 375𝑥 2 37) −𝑥 3 + 27 38) −𝑦 3 + 64 39) 64𝑥 3 + 27 40) 8𝑥 3 + 125 41) 𝑥 6 + 8 42) 𝑥 9 + 64

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