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Presentation transcript:

Warm Up

L6-3 Objective: Students will divide polynomials

Use long division to divide 125 by 3 2304 ÷ 7

Divide x2 + 2x – 30 by x – 5. Dividing Polynomials Additional Examples LESSON 6-3 Additional Examples Divide x2 + 2x – 30 by x – 5.

Determine whether x + 2 is a factor of each polynomial. Dividing Polynomials LESSON 6-3 Additional Examples Determine whether x + 2 is a factor of each polynomial. a. x2 + 10x + 16 b. x3 + 7x2 – 5x – 6 x2 + 5x – 15 x + 2 x3 + 7x2 – 5x – 6 x3 + 2x2 5x2 – 5x 5x2 + 10x –15x – 6 –15x – 30 24 x + 8 x + 2 x2 + 10x + 16 x2 + 2x 8x + 16 Since the remainder is zero, x + 2 is a factor of x2 + 10x + 16. Since the remainder = 0, x + 2 is not a factor of x3 + 7x2 – 5x – 6. /

Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3. Dividing Polynomials LESSON 6-3 Additional Examples Use synthetic division to divide 5x3 – 6x2 + 4x – 1 by x – 3. Step 1: Reverse the sign of the constant term in the divisor. Write the coefficients of the polynomial in standard form. – Write x 5x3 – 6x2 + 4x – 1 3 as 5 – 6 4 – 1 Step 2: Bring down the coefficient. Bring down the 5. 3 5 –6 4 –1 This begins the quotient. 5

Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9. Dividing Polynomials LESSON 6-3 Additional Examples Use synthetic division to find P(3) for P(x) = x4 – 2x3 + x – 9. By the Remainder Theorem, P(3) equals the remainder when P(x) is divided by x – 3. 3 1 –2 0 1 –9 3 3 9 30 1 1 3 10 21 The remainder is 21, so P(3) = 21.

Find P(4) is P(x)= 2x³ - 4x² + x – 5 Find P(-3), P(x)= x³ -5x² + 8

Homework L6-3 (p 330) #2-34e 42-46e