holt algebra 2 6-3 dividing polynomials use long division and synthetic division to divide...
TRANSCRIPT
Holt Algebra 2
6-3 Dividing Polynomials
Use long division and synthetic division to divide polynomials.
Objective
Holt Algebra 2
6-3 Dividing Polynomials
Polynomial long division is a method for dividing a polynomial by another polynomials of a lower degree. It is very similar to dividing numbers.
Holt Algebra 2
6-3 Dividing Polynomials
Divide using long division.
Example 1: Using Long Division to Divide a Polynomial
(–y2 + 2y3 + 25) ÷ (y – 3)
2y3 – y2 + 0y + 25
Step 1 Write the dividend in standard form, includingterms with a coefficient of 0.
Step 2 Write division in the same way you would when dividing numbers.
y – 3 2y3 – y2 + 0y + 25
Holt Algebra 2
6-3 Dividing Polynomials
Step 3 Divide.
2y2
–(2y3 – 6y2) 5y2 + 0y
+ 5y
–(5y2 – 15y)
15y + 25
–(15y – 45)
70
+ 15
Example 1 Continued
y – 3 2y3 – y2 + 0y + 25
Holt Algebra 2
6-3 Dividing Polynomials
Step 4 Write the final answer.
Example 1 Continued
–y2 + 2y3 + 25y – 3 = 2y2 + 5y + 15 +
70y – 3
Holt Algebra 2
6-3 Dividing PolynomialsDivide using long division.
(15x2 + 8x – 12) ÷ (3x + 1)
Holt Algebra 2
6-3 Dividing PolynomialsDivide using long division.
(x2 + 5x – 28) ÷ (x – 3)
Holt Algebra 2
6-3 Dividing Polynomials
Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 and a coefficient for any missing terms, and the divisor must be in the form (x – a).
Holt Algebra 2
6-3 Dividing Polynomials
Holt Algebra 2
6-3 Dividing Polynomials
Divide using synthetic division.
(3x4 – x3 + 5x – 1) ÷ (x + 2)
Step 1 Find a.
Use 0 for the coefficient of x2.
For (x + 2), a = –2.a = –2
Example 2B: Using Synthetic Division to Divide by a Linear Binomial
3 – 1 0 5 –1 –2
Step 2 Write the coefficients and a in the synthetic division format.
Holt Algebra 2
6-3 Dividing Polynomials
Example 2B Continued
Draw a box around the remainder, 45.
3 –1 0 5 –1 –2
Step 3 Bring down the first coefficient. Then multiply and add for each column.
–6
3 45
Step 4 Write the quotient.
3x3 – 7x2 + 14x – 23 +45
x + 2Write the remainder over the divisor.
46–2814
–2314–7
Holt Algebra 2
6-3 Dividing Polynomials
Divide using synthetic division.
(6x2 – 5x – 6) ÷ (x + 3)
Step 1 Find a.
Write the coefficients of 6x2 – 5x – 6.
For (x + 3), a = –3.a = –3
–3 6 –5 –6
Step 2 Write the coefficients and a in the synthetic division format.
Holt Algebra 2
6-3 Dividing Polynomials
Draw a box around the remainder, 63.
6 –5 –6 –3
Step 3 Bring down the first coefficient. Then multiply and add for each column.
–18
6 63
Step 4 Write the quotient.
6x – 23 +63
x + 3Write the remainder over the divisor.
–23
69
Holt Algebra 2
6-3 Dividing Polynomials
Divide using synthetic division.
(x2 – 3x – 18) ÷ (x – 6)
Holt Algebra 2
6-3 Dividing Polynomials
You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem.
Holt Algebra 2
6-3 Dividing Polynomials
Example 3A: Using Synthetic Substitution
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = 2x3 + 5x2 – x + 7 for x = 2.
Write the coefficients of the dividend. Use a = 2.
2 5 –1 7 2
4
2 41
P(2) = 41
Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7.P(2) = 2(2)3 + 5(2)2 – (2) + 7
P(2) = 41
3418
179
Holt Algebra 2
6-3 Dividing Polynomials
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = x3 + 3x2 + 4 for x = –3.
Holt Algebra 2
6-3 Dividing Polynomials
Use synthetic substitution to evaluate the polynomial for the given value.
P(x) = 5x2 + 9x + 3 for x = -2 .
Holt Algebra 2
6-3 Dividing Polynomials
Lesson Quiz
2. Divide by using synthetic division. (x3 – 3x + 5) ÷ (x + 2)
1. Divide by using long division. (8x3 + 6x2 + 7) ÷ (x + 2)
194; –4
3. Use synthetic substitution to evaluate P(x) = x3 + 3x2 – 6 for x = 5 and x = –1.
8x2 – 10x + 20 – 33
x + 2
x2 – 2x + 1 + 3
x + 2