University:
Montgomery College
Course:
MATH 050: Foundations of Algebra
Academic year:

2019
Views:

71

Pages:

2
Author:

Mia Hobbs

Simplifying Algebraic Expressions to Standard Polynomial Form: How might we simplify an expression like 2(x  1)(x  2)(x + 3) – (x2  4)(x + 2) + 5x3 + (x2  1)(x2 + 1) – 10(x + 1) to standard polynomial form? One approach* would be to break the problem into the following steps: 1. Multiply each term separately, keeping the parentheses: FOIL 2(x  1)(x  2)(x + 3) – FOIL (x2  FOIL 4)(x + 2) + 5x3 2(x2  2x  x + 2)(x + 3) + (x2  1)(x2 + Multiply 1) – 10x(x + 1) (x4 + x2  x2  1) (x3 + 2x2  x  8) (10x2 + 10x) 2. Combine like terms: 2(x2  3x + 2)(x + 3)  (x3 + 2x2  x  8) + (x4  1) – (10x2 + 10x) 3. If necessary, repeat steps 1 and 2, until all terms have been multiplied: Multiply each term by each term 2(x2  3x + 2)(x + 3)  (x3 + 2x2  x  8) 2(x3  3x2 + 2x + 3x2  9x + 6) + (x4  1) – (10x2 + 10x) These terms stay the same 2(x3  7x + 6) (2x3  14x + 12)  (x3 + 2x2  x  8) + (x4  1) – (10x2 + 10x) 3. When all terms have been multiplied, distribute the negative () signs: (2x3  14x + 12) + x3  2x2 x  8) + (x4  1) + (– 10x2 – 10x) *If you already have an approach that works for you, that’s great! Remember that there’s almost always more than one right way to approach a math problem. 4. Now drop the parentheses: 2x3  14x + 12  x3  2x2 x  8 + x4  1 – 10x2 – 10x 5. Find the term with the highest power of x. Find all terms with this power and combine. Move the result to the beginning: x4 + 2x3  14x + 12  x3  2x2 x  8  1 – 10x2 – 10x 6. Repeat for the next highest power of x: x4 + x3  14x + 12  2x2 x  8  1 – 10x2 – 10x 7. Continue until all like terms have been combined: x4 + x3  12x2  14x + 12 x  8  1 – 10x x4 + x3  12x2  20x + 12  8  1 x4 + x3  12x2  20x + 19 And we’re done! Exercises: Simplify the following expressions: 1. (x + 1)(x + 4)  (2x3  1) + (x2  1)(x  2) 2. 6x4  4(x  1)(x3 + 1)  5(x2  1)  x3  10 Answers: 1.x3  x2 + 4x + 7 2. 2x4 +3x3 +5x2  4x  1