1. Chapter 6 Review MTH 065 – Elementary Algebra

2. The Graph of f(x) = x 2 + bx + c vs. Solutions of x 2 + bx + c = 0 vs. Factorization of x 2 + bx + c 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.

3. The Graph of f(x) = x 2 + bx + c vs. Solutions of x 2 + bx + c = 0 vs. Factorization of x 2 + bx + c Three 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?

4. The Principle of Zero Products If ab = 0, then either a = 0, b = 0, or both. Using this Principle Solve: x(x – 5)(2x + 1) = 0 Solution: x = 0 x – 5 = 0  x = 5 2x + 1 = 0  x = –½ Therefore … x = 0, 5, –½

5. Factoring – Step 1 … always! Factor out common factors. Examples: Constant: 6x 2 – 12x – 21 = 3(2x 2 – 4x – 7) Variable: x 3 + 5x 2 – 2x = x(x 2 + 5x – 2) Both: –4x 4 + 32x 3 – 20x 2 = –4x 2 (x 2 – 8x + 5)

6. Factoring by Grouping (4 terms) x 3 – 2x 2 + 5x – 10 Make two groups of two terms. (x 3 – 2x 2 ) + (5x – 10) Factor out common factors from each group. x 2 (x – 2) + 5(x – 2) If the remaining binomials are identical, factor them out. (x – 2)(x 2 + 5) NOTE: Be careful with the sign of the third term when it is negative!

7. Factoring: x 2 + bx + c Leading coefficient = 1 Find 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.

8. Solving: x 2 + 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

9. Factoring: ax 2 + bx + c Leading coefficient > 1 Find two numbers m & n where … mn = ac m + n = b Rewrite the polynomial: ax 2 + 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.

10. Solving: ax 2 + 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

11. Leading Coefficient Negative Factor out –1 first, and then proceed as before. Example: –8x 2 + 10x + 3 = –[8x 2 – 10x – 3] = –[8x 2 – 12x + 2x – 3] = –[(8x 2 – 12x) + (2x – 3)] = –[4x(2x – 3) + 1(2x – 3)] = –(2x – 3)(4x + 1)

12. Factoring: Special Case #1 Perfect Square Polynomials ax 2 + bx + c If a is a perfect square … a = m 2 If c is a perfect square … c = n 2 If b = 2mn Then it factors as … If b is positive: (mx + n) 2 If b is negative: (mx – n) 2

13. Factoring: Special Case #2 Difference of two squares. ax 2 – c If a is a perfect square … a = m 2 If c is a perfect square … c = n 2 Then it factors as … (mx + n)(mx – n) NOTE: It must be subtraction! If it is addition, it will not factor.

14. Factoring Polynomials in Quadratic Form af(x) 2 + bf(x) + c Let y = f(x) (i.e. replace f(x) with y) ay 2 + 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.

15. 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