April 6, 2009 You need:textbook calculator No Fantastic Five warm ups this week. Take notes and/or read section Work together if you need help – teach each other!

10.5 Factoring x 2 + bx + c (a quadratic trinomial) If you can factor a polynomial in the form ax 2 + bx + c = 0, then you can find the roots to the parabola without graphing it. Remember roots are x-intercepts, where y = 0. Therefore, if you set each binomial equal to zero and solve for x, then you can find the roots. (x + 3)(x – 2) (x + 3) = 0(x + 2) = 0 x + 3 = 0x + 2 = 0 x = -3 x = -2

10.5 Factoring continued In 10.5, you will be factoring quadratic trinomials that have a leading coefficient of 1, meaning the equations will look like x 2 + bx + c (where a = 1). To factor a quadratic expression means to write it as a product (*) of two linear polynomials (x 1 ). (x + m) (x + n) To find the factors m and n, you are looking for two numbers that Multiply to equal the c in ax 2 + bx + c. Add up to equal the b in ax 2 + bx + c.

For example, to factor the polynomial x 2 + 9x List factors of 18: NOW, which factors ADD UP to 9? Your M and N factors are ___3 and 6____.

Replace M and N with your factors of 3 and 6: (x + m) (x + n) (x + 3) (x + 6)OR(x + 6) (x + 3) CHECK! Use the FOIL or Dist. Prop. to see if you get the original polynomial back (original = x 2 + 9x + 18). You should get x 2 + 6x + 3x + 18 (either method) which simplifies to x 2 + 9x + 18.

Finding the Final Answer to Factoring a Polynomial in x 2 + bx + c form: Once you find your two factors - - (x + 3)(x + 6), Set each factor equal to zero to find the roots. (x + 3) (x + 6) (x + 3) = 0(x + 6) = 0 x + 3 = 0x + 6 = 0 x = -3 x = -6 The ROOTS to the parabola are -3 and -6.

Confused? Read over pages , examples 1-6. Practice in class: p.607, #5-8 matching HOMEWORK: p.607, # 13, 16, 19, 22, 25, 29, 34, 41