Respuesta :
The series of operations for each case are listed below:
- GCF / GCF / GCF
- GCF / Grouping
- Quadratic trinomial
- GCF / Quadratic trinomial
- Difference of squares
- Difference of cubes / Quadratic trinomial
- Sum of cubes
- GCF / Quadratic trinomial
- GCF / Difference of squares
How to applying factor properties to simplify algebraic expressions
In algebra, factor properties are commonly used to solve certain forms of polynomials in a quick and efficient way and whose effectiveness is sustained on all definitions and theorems known in real algebra. In this problem, we should explain and show what factor properties are used in each case:
Case 1
5 · x · y³ + 10 · x² · y Given
5 · (x · y³ + 2 · x² · y) GCF
5 · x · (y³ + 2 · x · y) GCF
5 · x · y · (y² + 2 · x) GCF
Case 2
6 · z · x + 9 · x + 14 · z + 21 Given
3 · x · (z + 3) + 7 · (z + 3) GCF
(3 · x + 7) · (z + 3) Grouping
Case 3
a² + 2 · a - 63 Given
(a + 9) · (a - 7) Quadratic trinomial
Case 4
6 · z² + 5 · z - 4 Given
6 · [z² + (5 / 6) · z - 2 / 3] GCF
6 · (z - 1 / 2) · (z + 4 / 3) Quadratic trinomial
Case 5
81 · m² - 25 Given
(9 · m + 5) · (9 · m - 5) Difference of squares
Case 6
8 · x³ - 27 Given
(2 · x - 3) · (4 · x² + 6 · x + 9) Difference of cubes
4 · (2 · x - 3) · [x² + (3 / 2) · x + 9 / 4] Quadratic trinomial
Case 7
27 · b³ + 64 · z³ Given
(3 · b + 4 · z) · (9 · b² - 12 · b · z + 16 · z²) Sum of cubes
Case 8
2 · w³ - 28 · w² + 80 · w Given
2 · w · (w² - 14 · w + 40) GCF
2 · w · (w - 4) · (w - 10) Quadratic trinomial
Case 9
200 · a⁴ - 18 · b⁶ Given
2 · (100 · a⁴ - 9 · b⁶) GCF
2 · (10 · a² + 3 · b³) · (10 · a² - 3 · b³) Difference of squares
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