(a) long division (b)synthetic division Warm up 9/29   Dividing using (a) long division (b)synthetic division      x3+3x2-10x-1

Agenda: Warmup Notes 6.4 & 6.5 Go over ch 3 test if time Be seated before the bell rings Agenda: Warmup Notes 6.4 & 6.5 Go over ch 3 test if time DESK Warm-up (in your notes) homework

Notebook 6.4 Factor/6.5 Find roots 21) Factor and find roots HW Table of content 6.4 Factor/6.5 Find roots Page 19)Add, subtract, multiply polynomials 20) Dividing Polynomials 21) Factor and find roots 1 HW p.433(17-31 odd) p.442(15-22)

Factor 6.4 Factor/6.5 Find roots 1. f(x)= x2 – 7x 2. f(x)= x2 – 9

Factor by grouping 5. x3 – x2 – 25x + 25. (x3 – x2) + (–25x + 25) Group terms. x2(x – 1) – 25(x – 1) Factor GCF from each group. (x – 1)(x2 – 25) Factor out the common binomial (x – 1)(x – 5)(x + 5) Factor the difference of squares.

You try! 6. Factor: x3 – 2x2 – 9x + 18. (x3 – 2x2) + (–9x + 18)

(2x3 + x2) + (8x + 4) x2(2x + 1) + 4(2x + 1) (2x + 1)(x2 + 4) Check It Out! 7. Factor: 2x3 + x2 + 8x + 4. (2x3 + x2) + (8x + 4) x2(2x + 1) + 4(2x + 1) (2x + 1)(x2 + 4) (2x + 1)(x2 + 4)

8. (x3 + 27) 9. x 6 - 8 (x2)3 - (2)3 (x3 + 33) ( - )[( )2 +  + ( )2] x2 2 x2 z 2 2 (x + 3)(x2 – x  3 + 32) (x2 - 2 )(x4– 2x + 4) (x + 3)(x2 – 3x + 9)

10. 125d3 – 8 (5d)3 – 23 (5d – 2)[(5d)2 + 5d  2 + 22] (5d – 2)(25d2 + 10d + 4)

2x2(x3 – 8) 2x2(x3 – 23) 2x2(x – 2)(x2 + x  2 + 22)

   

Roots of a Polynomial Equation x-intercept, where graph crosses x-axis How to “solve” a polynomial equation: Set the equation equal to 0 Factor the Equation Set each factor equal to 0 and solve for roots!

Find the roots of the equation by factoring. 14. x2 – 4x = –4 x2 – 4x + 4 = 0 (x – 2)(x – 2) = 0 x – 2 = 0 or x – 2 = 0 x = 2 or x = 2

Solve the polynomial equation by factoring. 15. 4x6 + 4x5 – 24x4 = 0 4x4(x2 + x – 6) = 0 Factor out the GCF, 4x4. 4x4(x + 3)(x – 2) = 0 Factor the quadratic. 4x4 = 0 or (x + 3) = 0 or (x – 2) = 0 Set each factor equal to 0. x = 0, x = –3, x = 2 Solve for x. The roots are 0, –3, and 2.

Determine whether the given binomial is a factor of the polynomial P(x). 16. (x + 1); (x2 – 3x + 1) 17. (x + 2); (3x4 + 6x3 – 5x – 10) Find P(–1) by synthetic substitution. Find P(–2) by synthetic substitution. –1 1 –3 1 –2 3 6 0 –5 –10 –1 4 –6 10 1 –4 5 3 –5 P(–1) = 5 P(–2) = 0, so (x + 2) is a factor of P(x) = 3x4 + 6x3 – 5x – 10. P(–1) ≠ 0, so (x + 1) is not a factor of P(x) = x2 – 3x + 1.

18. x3 + 3x2 – 4x – 12 = 0 Factors of –12: ±1, ±2, ±3, ±4, ±6, ±12 Steps: Identify possible roots Test to find one that makes the equation true That is a factor, divide to see what’s left Factor more if necessary or use quadratic formula if possible 18. x3 + 3x2 – 4x – 12 = 0 Factors of –12: ±1, ±2, ±3, ±4, ±6, ±12

19. Identify all the real roots of 2x3 – 3x2 –10x – 4 = 0. Steps: Identify possible roots Test to find one that makes the equation true That is a factor, divide to see what’s left Factor more if necessary or use quadratic formula if possible

20. Identify all the real roots of 2x3 – 42x+40 = 0. Steps: Identify possible roots Test to find one that makes the equation true That is a factor, divide to see what’s left Factor more if necessary or use quadratic formula if possible

19. Identify all the real roots of 4x5 – 8x3+4x = 0. Steps: Identify possible roots Test to find one that makes the equation true That is a factor, divide to see what’s left Factor more if necessary or use quadratic formula if possible

Divide the polynomial by 3, then find P(2) by synthetic substitution. Check It Out! Example 1 Determine whether the given binomial is a factor of the polynomial P(x). a. (x + 2); (4x2 – 2x + 5) b. (3x – 6); (3x4 – 6x3 + 6x2 + 3x – 30) Find P(–2) by synthetic substitution. Divide the polynomial by 3, then find P(2) by synthetic substitution. –2 4 –2 5 –8 20 2 1 –2 2 1 –10 4 –10 25 2 4 10 P(–2) = 25 1 2 5 P(–2) ≠ 0, so (x + 2) is not a factor of P(x) = 4x2 – 2x + 5. P(2) = 0, so (3x – 6) is a factor of P(x) = 3x4 – 6x3 + 6x2 + 3x – 30.

 

Solve the polynomial equation by factoring. 2x6 – 10x5 – 12x4 = 0 2x4(x2 – 5x – 6) = 0 Factor out the GCF, 2x4. 2x4(x – 6)(x + 1) = 0 Factor the quadratic. 2x4 = 0 or (x – 6) = 0 or (x + 1) = 0 Set each factor equal to 0. x = 0, x = 6, x = –1 Solve for x. The roots are 0, 6, and –1.

Solve the polynomial equation by factoring. x4 + 25 = 26x2 x4 – 26 x2 + 25 = 0 Set the equation equal to 0. Factor the trinomial in quadratic form. (x2 – 25)(x2 – 1) = 0 (x – 5)(x + 5)(x – 1)(x + 1) Factor the difference of two squares. x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0 x = 5, x = –5, x = 1 or x = –1 Solve for x. The roots are 5, –5, 1, and –1.