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14

MULTIPLYING OUT

THE DISTRIBUTIVE RULE

 

m(a + b) = ma + mb

"To multiply a sum by a number,
 multiply each term of the sum."

That is called the distributive rule.  m multiplies a, then it multiplies b. We say that we have "distributed" m to a and b.

(Compare Lesson 9 of Arithmetic.)

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
 
3x4· −5x = −15x5,  The Rule of Signs
 
3x4· 1 = 3x4.  

Whenever a power of x is the multiplier: 3x4 is the multiplier; and it multiplies powers of x, then always add the exponents.

Problem 1.    −1(ab + cd)

What will be the effect of multiplying by −1?

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

Every sign will change.  −1(ab + cd) = −a + bc + d

It follows, then, that we may change all the signs on both sides of an equation.

This equation

  x + ab  =  c
 
is equivalent to this one:
 
  xa + 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)   3x2(4x3 − 3x2 + 5x − 8) = 12x5 − 9x4 + 15x3 − 24x2  
 
  f)   −5x4(x3 − 4x2 + 2x − 6) = −5x7 + 20x6 − 10x5 + 30x4  

g)   2xy(x2 − 3xy + y2 = 2x3y − 6x2y2 + 2xy3

h)   −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(5xy)   =   8x + 10y + 15x − 3y
 
    =   23x + 7y
  b)   4(2x − 1) − 5(x − 2)   =   8x − 4 − 5x + 10
 
    =   3x + 6
  c)   3x(3x − 2y) − 2y(xy)   =   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 + ab2ba2 + 2ab2b3
 
  =  a3 − 3a2b + 3ab2b3

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  (pq)(xy + z).   Observe the Rule of Signs (Lesson 4).

(pq)(xy + z) = pxpy + pz  −  qx + qyqz

Example 3.   Multiply out  (x2)(x + 3).  Simplify by adding the like terms.

Solution.   First distribute x, then distribute −2:

(x2)(x + 3) = x· x + x· 3  − 2· x2· 3
 
  = x2 + 3x2x − 6
 
  = x2 + x − 6.

The student should not have to write the first line above, but should be able to write the second line --

x2 + 3x2x − 6

-- immediately.

(In Lesson 16, we will see a more skillful way to arrive at the answer.)

Vertical multiplication, as done in arithmetic, should be avoided. It is all right, perhaps, when algebra is a terminal course and the teacher is interested only in getting students to pass a text, skill not being an issue. But for students who will continue on to calculus, it will be a habit difficult to overcome. The student will never see vertical multiplication in any calculus text, or any physics text, and the student conditioned to seeing vertical multiplication will find those texts difficult to read. The teacher who allows vertical multiplication does not do those students any favor.

The same can be said for teaching any method for solving equations other than transposing, and the general method of inverse operations.

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 + 8xx − 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
 
  = x3x2  − 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

Note:  The effect of multiplying by x is simply to increase each exponent by 1.

  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  x2x − 1
 
  = x4 − 1

Note:  Multiplication by −1 simply changes the signs.

Next Lesson:  Common Factor


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