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# 3:6 Solving EquationsbyFactoring

last edited by 3 years, 11 months ago

# LESSON 3.6  Solving Equations by Factoring

## Do you remember multiplyingwith ZERO: 5(0)=0 or 0(-9)=0 or 0(0)=0.

That's easy!!

It  is simply arithmetic,  but it is important in our next use of factoring.  I hope you recall factoring.

We have had several lessons on that topic.

Factoring is writing an algebra expression as MULTIPLICATION.

When we multiply with zero our answer is always zero.

Do you recall 0 times 0=0   or that 5 times 0 = 0 ?

No other number has that quality.

### Right?

Thus (x-5)(x+9)=ZERO means that    x- 5=0 or that the x+9=0.

NOW  Solve the above two algebra equations:

x-5=0

so x= +5

x+9=0

so x= -9.

### It too will make our equation true.

Solve (x-5)(x+9)=0

means x= 5 or x=-9 .

These are called solutions.

## Do you know where I got this quadratic trinomial?

I hope so.  If not then multiply (x-5)(x+9).   YOU must know how; some call it FOIL  others call it the distributive property.

(x-5) times (x+9) =  x2+9x-5x-45 simplifies to x2+4x -45.

If I take the trinomial x2+4x-45 and substitute -9 for each x then

I will have (-9)2+4(-9)-45

=81-36-45 which is 0.

If I take the trinomial x2+4x - 45 and substitute a 5 for each x then

I will have  (5)2+4(5)-45

=25+20-45 which is 0.

## We call those numbers the solutionto x2+4x-45.

Let us solve some polynomial equations.   Here are the steps that we used above.
We will  :

### 3.  solve each of the  simple equations for x.

( WE will have several linear equations to solve.)

EXAMPLE:   x2 -17x +72 = 0

becomes (x- 9)(x -8) =0.

WE Factored.

we have x-9 =0 and x-8 =0. Now solve for x.

Solutions are x=9 and x=8.

You try these.   Write these in your notebook so that you have good examples to follow.

t2  - 12 = 4t

We must begin by rewriting.

18m2 - 8 = 0 Note the common factor of 2.

2(9m2 - 4=0

t4- 13t2 + 36 = 0

it is a quartic.

t2  - 4t  -12 = 0

2(3m -2)(3m+2) = 0

The 2 cannot=0, but the 3m-2 and the 3m+2 can =0.

Use Algebra 1 to solve for m.

(t2-9) (t2-4)  =0

(t  -6)(t +2) = 0

3m -2 =0  or 3m+ 2 =0

t2-9   = 0  or t2-4  = 0

t  - 6 =0  or t+2 = 0

3m = 2   or 3m = -2

(t - 3)(t+3)=0    or    (t - 2)(t+2)=0

t= 6  or t= -2

m = 2/3  or m = - 2/3

t=3  or    t= -3    or    t=2  or t= -2

## 4 solutions!

The highest exponent will tell you!!

I have created two videos for working "word problems" with factoring.

Please visit my web site www.mathinabox.com and click the tab "Lessons / Videos".

You will see the two videos listed  at the bottom of the middle list.

****************************************************************************************

EXAMPLE with WORD problem:

### Its area is 240 ft2.

Find the equation to express its area in terms of W,  then solve to find its width and length.

W= width, W+8=length          See above that it state the length is 8 more than the width.

### Area = length times width

240  = (W+8) times W

This becomes: 240 = W2 + 8w

This is a quadratic and we can find the solutions by

rewriting=0,

factoring and then solving.

0 = W2 + 8W -240  Be sure you know why the 240 is negative now.

0=(W +20)(W - 12)

0=W+20 or W-12=0

W=-20 or W=12

Since the width of a rectangle can not be negative the width must be 12 feet.

Length is 8 more than width=20 feet.

1.  Five times an integer plus three times the square of the integer is twenty-two. Find the integer.

There are many problems that can be set  up and solved using factoring.

The world of science, economics, engineering, architecture, electricity are only a few to consider.

Remember multiplying with 0 yields 0 .

If 2x (x+4) = 0 then one of the factors has to be zero.

To multiply and get zero for an answer always means one or both factors we multiplied with was equal to zero.

Therefore either 2x =0 or x+4 = 0 or both are equal to zero.

Solve these equations:

2x =0        or      x+4 = 0

x = 0        or      x = -4.

MORE EXAMPLES

Find the solutions.

Example :  Easy one, since it is already equal to zero and factored .

2x(x+4)=0    so      2x=0  or x+4 =0

x=0  or x=- 4 are the solutions for 2x(x+4)=0 .

Check them by substituting -4 for the x in the original problem.

2x(x+4)

2(-4)[-4+4] =-8=0

Yes that is what we wanted, zero.

You check the x=0.

Example: 2x3- 9x2=0

or I could write it as 2x3=9x2

Either way I MUST follow these steps:

1.  rewrite it so that it =zero,

2.  then factor,

3.  then solve each EQUATION.

EXAMPLE:     x4- 9x2=0  There is a common factor  of x2

(I hope you recall  always do COMMON FACTOING first IF THERE IS ANY)

So this can be:

x2 (x2-9) =0    NOW FACTOR THE x2 - 9

### x2= 0  or   x-3=0  or        x+3 =0

There are 3 equations to solve.

The x2 = 0 means x = 0.

The x - 3 = 0 means x= 3 .

The  x+3 = 0 means x = - 3.

Our solutions are x=0 or x=3 or x= -3.

Can you check these by substitution into the original problem?  Let x be a 0 in the equation.   Does it make the equation true?

Let x be a 3  or -3 in the equation.   Does it make the equation true?   It should.

Solve by factoring.

Please email these to me at sojohnsey@gmail.com   Write Factoring class 3.6 in the subject line of the email.

If you can not factor them then send what you have of each problem.

You should rewrite it first(if needed) then factor and then solve for the x's.

## Although I have this in 3 columns please send your work in only one long column.

 1.  Completely FACTOR: x4 - 27x 2.  SOLVE:   t2 - 16 = 0 3.  SOLVE:  x2 + 8x + 16 = 0 4.  SOLVE:  p2 + 3p = 4 5.  SOLVE:  v2 + 12 = 8v 6. SOLVE:   3x2 - 7x - 6 = 0 7.  SOLVE:  3x2 -10x -8= 0 8. SOLVE:   x - 9=x( x - 5)         HINT:   Multiply these out and then rewrite =0 so that they look like #3 or #7 above. 9.  SOLVE:  t + 3 = t( t + 3)   HINT:   Multiply these out and then rewrite =0  so that they look like #3 or #7 above. 10.  SOLVE:   u2 -3u+9=( u - 1)(2 u + 3)    HINT:   Multiply these out and then rewrite so that they look like #3 or #7 above.

Do Assignment  3.6 in SECTION 1.