14
MULTIPLYING OUT
THE DISTRIBUTIVE RULE
m(a + b) = ma + mb
"To multiply a sum, multiply each term of the sum."
That is called the distributive rule. m multiplies a, then it multiplies b. We say that m is "distributed" to a and b.
Example 1. 2(x + y + z) = 2x + 2y + 2z.
We have distributed 2 to x, y, and z. We have "multiplied out."
Example 2. 3x4(x² − 5x + 1) = 3x6 − 15x5 + 3x4
That is,
| 3x4· x² | = | 3x6, Rule 1 of exponents (Lesson 13) |
| 3x4· −5x | = | −15x5 |
| 3x4· 1 | = | 3x4 |
Problem 1. −1(a − b + c − d)
What will be the effect of multiplying by −1?
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Every sign will change. −1(a − b + c − d) = −a + b − c + d
Therefore in an equation, we may change all the signs on both sides. This equation
| −x + a − b | = | c | |
| implies this one: | |||
| x − a + b | = | −c. | |
Theoretically, we have multiplied both sides by −1.
Problem 2. Multiply out.
| a) | 5(x + 4) = 5x + 20 | b) | 5(x − 4) = 5x − 20 | ||
| c) | x(x + 1) = x² + x | d) | 2x(3x² + 5x − 6) = 6x3 + 10x² − 12x | ||
| e) | −5x4(x3 − 4x² + 2x − 6) = −5x7 + 20x6 − 10x5 + 30x4 | ||||
f) 2xy(x² − 3xy + y²) = 2x3y − 6x²y² + 2xy3
g) −4xy²(x3y − 6xy² − 2x + 3y + 1)
= −4x4y3 + 24x²y4 + 8x²y² − 12xy3 − 4xy²
Problem 3. Multiply out and simplify, that is, add the like terms.
| a) 2(4x + 5y) + 3(5x − y) | = | 8x + 10y + 15x − 3y |
| = | 23x + 7y | |
| b) 4(2x − 1) − 5(x − 2) | = | 8x − 4 − 5x + 10 |
| = | 3x + 6 | |
| c) 3x(3x − 2y) − 2y(x − y) | = | 9x² − 6xy − 2yx + 2y² |
| = | 9x² − 8xy + 2y² | |
| d) | x(x² − 10x + 25) − 5(x² − 10x + 25) |
| = x3 − 10x² + 25x − 5x² + 50x − 125 | |
| = x3 − 15x² + 75x − 125 | |
| e) | a(a² − 2ab + b²) − b(a² − 2ab + b²) |
| = a3 − 2a²b + ab² − ba² + 2ab² − b3 | |
| = a3 − 3a²b + 3ab² − b3 | |
A sum by a sum
(a + b + c)(x + y + z)
First distribute a to x, y, and z.
Then distribute b.
Then distribute c.
(a + b + c)(x + y + z)
= ax + ay + az + bx + by + bz + cx + cy + cz
Problem 4. Multiply (p − q)(x − y + z). Observe the Rule of Signs (Lesson 4).
(p − q)(x − y + z) = px − py + pz − qx + qy − qz
Example 3. Multiply out (x − 2)(x + 3). Simplify by adding the like terms.
Solution. First distribute x, then distribute −2:
| (x − 2)(x + 3) | = | x· x + x· 3 − 2· x − 2· 3 |
| = | x² + 3x − 2x − 6 | |
| = | x² + x − 6 | |
The student should not have to write the first line, but should be able to write the second line --
x² + 3x − 2x − 6
-- immediately.
Problem 5. Multiply out. Always simplify by adding the like terms.
| a) (x + 5)(x + 2) | = | x² + 2x + 5x + 10 |
| = | x² + 7x + 10 | |
| b) (x + 5)(x − 2) | = | x² − 2x + 5x − 10 |
| = | x² + 3x − 10 | |
| c) (x − 5)(x − 2) | = | x² − 2x − 5x + 10 |
| = | x² − 7x + 10 | |
| d) (2x − 1)(x + 4) | = | 2x² + 8x − x − 4 |
| = | 2x² + 7x − 4 | |
| e) (3x + 2)(4x − 5) | = | 12x² − 15x + 8x − 10 |
| = | 12x² − 7x − 10 | |
| f) (5x − 1)² | = | (5x − 1) (5x − 1) |
| = | 25x² − 5x − 5x + 1 | |
| = | 25x² − 10x + 1 | |
| g) (6x + 1)(6x − 1) | = | 36x² − 6x + 6x − 1 |
| = | 36x² − 1 | |
| Example 4. (x − 4)(x² + 3x − 10) | = | x3 + 3x² | − 10x |
| − 4x² | − 12x + 40 | ||
| = | x3 − x² | − 22x + 40 | |
Notice: Upon distributing −4, we have anticipated the like terms by aligning them. However, that is not strictly necessary.
Problem 6. Multiply out.
| a) (x + 2)(x² + 4x − 5) | = | x3 + 4x² | − 5x |
| + 2x² | + 8x − 10 | ||
| = | x3 + 6x² | + 3x − 10 | |
| b) (x − 3)(x² − 6x + 9) | = | x3 − 6x² | + 9x |
| − 3x² | + 18x − 27 | ||
| = | x3 − 9x² | + 27x − 27 | |
| c) (3x − 4)(x² − 7x − 2) | = | 3x3 − 21x² | − 6x |
| − 4x² | + 28x + 8 | ||
| = | 3x3 − 25x² | + 22x + 8 | |
| d) (x − 1)(x3 + x² + x + 1) | = | x4 + x3 | + x² + x |
| − x3 | − x² − x − 1 | ||
| = | x4 − 1 | ||
Note: The effect of multiplying by x is to increase each exponent by 1; the effect of multiplying by −1 is to change each sign.
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