Chapter 6Review

MTH 065 – Elementary Algebra

The Graph of f(x) = x2 + bx + cvs.

Solutions of x2 + bx + c = 0vs.

Factorization of x2 + bx + c

x

y

The graph is a parabola.

The solutions are the x-intercepts.

The factors are (x – m)(x – n), where m and n are the x-intercepts.

The Graph of f(x) = x2 + bx + cvs.

Solutions of x2 + bx + c = 0vs.

Factorization of x2 + bx + cThree possibilities …

x = 3, –1

(x – 3)(x + 1)

x = –2

(x + 2)2

No Solution

Does not Factor

What if an x-intercept is the origin?

The Principle of Zero ProductsIf ab = 0, then either a = 0, b = 0, or both.

Using this Principle

Solve: x(x – 5)(2x + 1) = 0

Solution:x = 0x – 5 = 0 x = 52x + 1 = 0 x = –½

Therefore …

x = 0, 5, –½

Factoring – Step 1 … always!

Factor out common factors.Examples:

Constant: 6x2 – 12x – 21 = 3(2x2 – 4x – 7)

Variable:x3 + 5x2 – 2x = x(x2 + 5x – 2)

Both: –4x4 + 32x3 – 20x2 = –4x2(x2 – 8x + 5)

Factoring by Grouping (4 terms)

x3 – 2x2 + 5x – 10

Make two groups of two terms.

(x3 – 2x2) + (5x – 10)

Factor out common factors from each group.

x2(x – 2) + 5(x – 2)

If the remaining binomials are identical, factor them out.

(x – 2)(x2 + 5)

NOTE: Be careful with the sign of the third term when it is negative!

Factoring: x2 + bx + cLeading coefficient = 1Find two numbers m & n where …

mn = c m + n = b

Then the factors are … (x + m)(x + n)

Signs …If c is positive, then both numbers have the

same sign as b.If c is negative, then the “larger” one has the

same sign as b and the other one has the opposite sign.

Solving: x2 + bx + c = 0

All terms to the left or right (one side MUST be 0).

Leading coefficient = 1

Factor the polynomial …

(x + m)(x + n) = 0

Set each factor equal to 0 and solve.

i.e. Solutions are: x = – m, –n

Factoring: ax2 + bx + cLeading coefficient > 1Find two numbers m & n where …

mn = ac m + n = b

Rewrite the polynomial: ax2 + mx + nx + c

Factor by grouping (be careful when nx is negative).

Signs …If ac is positive, then both numbers have the same

sign as b.If ac is negative, then the “larger” one has the

same sign as b and the other one has the opposite sign.

Solving: ax2 + bx + c = 0

All terms to the left or right (one side MUST be 0).

Leading coefficient > 1

Factor the polynomial (ac method) …

(rx + s)(tx + u) = 0

Set each factor equal to 0 and solve.

i.e. Solutions are: x = – s/r, –u/t

Leading Coefficient Negative

Factor out –1 first, and then proceed as before.

Example: –8x2 + 10x + 3 = –[8x2 – 10x – 3] = –[8x2 – 12x + 2x – 3] = –[(8x2 – 12x) + (2x – 3)] = –[4x(2x – 3) + 1(2x – 3)] = –(2x – 3)(4x + 1)

Factoring: Special Case #1

Perfect Square Polynomials

ax2 + bx + cIf a is a perfect square … a = m2

If c is a perfect square … c = n2

If b = 2mn

Then it factors as …If b is positive: (mx + n)2

If b is negative: (mx – n)2

Factoring: Special Case #2

Difference of two squares.

ax2 – cIf a is a perfect square … a = m2

If c is a perfect square … c = n2

Then it factors as …(mx + n)(mx – n)

NOTE: It must be subtraction! If it is addition, it will not factor.

Factoring Polynomials in Quadratic Form

af(x) 2 + bf(x) + c

Let y = f(x) (i.e. replace f(x) with y)

ay2 + by + c

Factor …

(my + n)(ry + s)

Substitute back (i.e. replace y with f(x)

(mf(x) + n)(rf(x) + s)

Simplify … if possible.

Applications …

1. Familiarize - read & summarize

1.5 Estimate - approximation

2. Translate - equation

3. Carry out - solve & answer

4. Check - reasonable

5. State - answer w/ units