1. MTH 065 Elementary Algebra II Chapter 6 – Polynomial Factorizations and Equations Section 6.2 – Equations Containing Trinomials of the Type x 2 + bx + c

2. Review: Multiplying Similar Binomials Consider binomials of the form (x + k) and (x – k). In the product … Where did the –18 come from? Where did the –3 come from? Why is the coefficient of x 2 equal to 1?

3. Review: Multiplying Similar Binomials FOIL First – Outside – Inside - Last F L I O

4. Factoring: The reverse of multiplying. If the product (x + 3)(x – 6) is x 2 – 3x – 18, then … The factorization of x 2 – 3x – 18 is (x + 3)(x – 6) Factorization is unique! Only one way to “completely” factor a polynomial. Commutative equivalences are not considered different factorizations.

5. Factoring: x 2 + bx + c Method x 2 + bx + c = (x + m)(x + n) mn = c m + n = b Signs of m & n c > 0 & b > 0  both positive c > 0 & b < 0  both negative c 0  “larger” is positive & “smaller is negative c < 0 & b < 0  “larger” is negative & “smaller” is positive b & c are integers

6. Factoring: x 2 + bx + c Examples Warning: Not all polynomials are factorable using integers.

7. One more reminder … 1 st Step of Factoring Always factor out the common factor first (if there is one). 3x 2 + 6x – 9 = 3(x 2 + 2x – 3) = 3(x + 3)(x – 1) If the leading coefficient is negative, factor out a -1. –x 2 – 7x + 2 = –(x 2 + 7x – 2) –5x 3 + 10x 2 – 5x = –5x(x 2 – 2x + 1) = -5x(x – 1)(x – 1)

8. Equations of the form … x 2 + bx + c = 0 1. Put the equation into standard form. x 2 – 5 = 3 – 2x x 2 + 2x – 8 = 0 2. Factor the polynomial (if possible). (x – 2)(x + 4) = 0 3. Set each factor equal to zero. x – 2 = 0 & x + 4 = 0 4. Solve the simpler equations. x = 2, –4 5. Check using the original equation.

9. Equations of the form … x 2 + bx + c = 0 More examples …

10. The relationship between … f(x) = x 2 + bx + c and x 2 + bx + c = 0 The graph of the function crosses the x-axis at the solutions of the equation.

11. The relationship between … f(x) = x 2 + bx + c and x 2 + bx + c = 0 The graph of the function crosses the x-axis at the solutions of the equation. Two Integer Solutions (i.e. factorable) f(x) = x 2 + x – 6 Solutions of x 2 + x – 6 = 0 are x = 2, –3 and x 2 + x – 6 = (x – 2)(x + 3)

12. The relationship between … f(x) = x 2 + bx + c and x 2 + bx + c = 0 The graph of the function crosses the x-axis at the solutions of the equation. One Integer Solution (i.e. factorable) f(x) = – x 2 + 2x – 1 Solution of –x 2 + 2x – 1 = 0 is x = 1 and –x 2 + 2x – 1 = –(x – 1) 2

13. The relationship between … f(x) = x 2 + bx + c and x 2 + bx + c = 0 The graph of the function crosses the x-axis at the solutions of the equation. Non-Integer Solutions (i.e. not factorable) f(x) = x 2 + x – 3 Solutions of x 2 + x – 3 = 0 are x = ?????? These will be solved in the next chapter.

14. The relationship between … f(x) = x 2 + bx + c and x 2 + bx + c = 0 The graph of the function crosses the x-axis at the solutions of the equation. No solutions (i.e. not factorable) f(x) = –x 2 + 2x – 3 Solutions of –x 2 + 2x – 3 = 0 do not exist! These will be considered in the next chapter.

15. One more problem … What functions are graphed below? f(x) = (x – 1)(x – 4) or f(x) = x 2 – 5x + 4 f(x) = x(x + 3) or f(x) = x 2 + 3x