Step-by-step explanation:

. Factoring and solving equations

- A. Factor1. Factor 3x2 + 6x if possible.

Look for monomial (single-term) factors first; 3 is a factor of both 3x2

and 6x and so is x . Factor them out to get

3x2 + 6x = 3(x2 + 2x1 = 3x(x+ 2) .

2. Factor x2 + x - 6 if possible.

Here we have no common monomial factors. To get the x2 term

we'll have the form (x +-)(x +-) . Since

(x+A)(x+B) = x2 + (A+B)x + AB ,

we need two numbers A and B whose sum is 1 and whose product is

-6 . Integer possibilities that will give a product of -6 are

-6 and 1, 6 and -1, -3 and 2, 3 and -2.

The only pair whose sum is 1 is (3 and -2) , so the factorization is

x2 + x - 6 = (x+3)(x-2) .

3. Factor 4x2 - 3x - 10 if possible.

Because of the 4x2 term the factored form wli be either

(4x+A)(x +B) or (2x+A)(2x+B) . Because of the -10 the integer possibilities for the pair A, B are

10 and -1 , -10 and 1 , 5 and -2 . -5 and 2 , PLUS each of

these in reversed order.

Check the various possibilities by trial and error. It may help to write

out the expansions

(4x + A)(x+ B) = 4x2 + (4B+A)x + A8

1 trying to get -3 here

(2x+A)(2x+B) = 4x2 + (2B+ 2A)x + AB

Trial and error gives the factorization 4x2 - 3x - 10 - (4x+5)(x- 2) .

4. Difference of two squares. Since (A + B)(A - B) = - B~ , any

expression of the form A' - B' can be factored. Note that A and B

might be anything at all.

Examples: 9x2 - 16 = (3x1' - 4' = (3x +4)(3x - 4)

x2 - 29 = x2 - (my)* = (x+ JTY)(x- my)

For any of the above examples one could also USE the  

In the factorization

-

ax2 + bx + c = a(x-Ah-B) ,

the numbers A and B are given by

A,B = -

2a

If the 'discriminant" b2 - 4ac is negative, the polynomial

cannot be factored over the real numbers (e.g. consider x2 + 1).

In EXAMPLE 2 above, a = 1, b = 1, c = -6 , so

A,B =

2 'l tE3 = '

= 2,-3 , so x +x-6 - (x-2)(~+3). 2 2

5. Factor x3 + 3x2 - 4 if possible.

If plugging x = a into a polynomial yields -zero, then the

polynomial has (x - a) as a factor.

We'll use this fact to try to find factors of x3 + 3x2 - 4 . We look for

factors (x-a) by plugging in various possible a's , choosing those that

are factors of -4 . Try plugging x 1-12 2 4, -4 into

x3+3x2-4. Findthat x-I gives 13+3.12-4 -0. So x-1 isa

factor of x3 + 3x2 - 4 . To factor it out, perform long division:

x2 + 4x + 4 Thus

x-~ lx3 + sx2+ Ox - ~3 + 3x2 - 4 = (x-I)(x~ + 4x + 4).

x3 - .2 But x2 + 4x + 4 can be

4x2 + Ox - 4 factored further as in the

4x2 - 4x examples above;

4x - 4

-

0

we finally get x3 + 3x2 - 4 = (x- l)(x+ 2)(x+2) = (x- 1)(~+2)~ .

s IIA Factor the following polynomials.

1. x2 + 8X + 15 2. 4x2 - 25  

3. 49 - 13y - 12 4. x3 + 2x2 - x - 2

5. 4z2 + 42 - 8 6. a2 + 3a + 2

7. Simplify by factoring 3x 2 + 3x - 18

numerator and denominator: 2 4x - 3x - 10

B. Solvina eauatlons

1. Linear or first-degree equations: involving x but not x2 or any

other power of x . Collect x-terms on one side, constant terms on the

other.

ExamDle x+3=7x-4

x + (-7x1 = -4 + (-3)

-6x = -7

x = 7/6

2. Quadratic equations: involving x2 but no higher power of x . These are solved by factoring and/or use of the quadratic formula:

The equation ax2 + bx + c = 0 (a - 0)

has solutions x =

2a

If b2 - 4ac is negative, the equation has no real solutions.

Exam& Solve x2-2x-3=0 for x.

Method: Factoring. x2 - 2x - 3 = (x- 3)(x+1) = 0 . Since a product of two numbers is zero if and only if one of the two

numbers is zero, we must have

x - 3 = 0 or x + I = 0 . So the solutions are x = 3. -1 .