Five-Minute Check (over Lesson 5–1)
CCSS
Then/Now
New Vocabulary
Example 1: Divide a Polynomial by a Monomial
Example 2: Division Algorithm
Example 3: Standardized Test Example: Divide Polynomials
Key Concept: Synthetic Division
Example 4: Synthetic Division
Example 5: Divisor with First Coeffecient Other than 1
Over Lesson 5–1
Simplify b2 ● b5 ● b3.
A. b5
B. b8
C. b10
D. b30
Over Lesson 5–1
A.
B.
C.
D.
Over Lesson 5–1
Simplify (10a2 – 6ab + b2) – (5a2 – 2b2).
A. 15a2 + 8ab + 3b2
B. 10a2 – 6ab – b2
C. 5a2 + 6ab – 3b2
D. 5a2 – 6ab + 3b2
Over Lesson 5–1
Simplify 7w(2w2 + 8w – 5).
A. 14w3 + 56w2 – 35w
B. 14w2 + 15w – 35
C. 9w2 + 15w – 12
D. 2w2 + 15w – 5
Over Lesson 5–1
State the degree of 6xy2 – 12x3y2 + y4 – 26.
A. 11
B. 7
C. 5
D. 4
Over Lesson 5–1
Find the product of 3y(2y2 – 1)(y + 4).
A. 18y5 + 72y4 – 9y3 – 36y2
B. 6y4 + 24y3 – 3y2 – 12y
C. –18y3 – 3y2 + 12y
D. 6y3 – 2y + 4
Content Standards
A.APR.6 Rewrite simple rational expressions
in different forms; write a(x)/b(x) in the form
q(x) + r(x)/b(x), where a(x), b(x), q(x), and
r(x) are polynomials with the degree of r(x)
less than the degree of b(x), using
inspection, long division, or, for the more
complicated examples, a computer algebra
system.
Mathematical Practices
6 Attend to precision.
You divided monomials.
• Divide polynomials using long division.
• Divide polynomials using synthetic division.
• synthetic division
Divide a Polynomial by a Monomial
Sum of quotients
Divide.
= a – 3b2 + 2a2b3
a1 – 1 = a0 or 1 and b1 – 1 = b0 or 1
Answer: a – 3b2 + 2a2b3
A. 2x3y – 3x5y2
B. 1 + 2x3y – 3x5y2
C. 6x4y2 + 9x7y3 – 6x9y4
D. 1 + 2x7y3 – 3x9y4
Division Algorithm
Use long division to find (x2 – 2x – 15) ÷ (x – 5).
x(x – 5) = x2 – 5x
–2x – (–5x) = 3x
3(x – 5) = 3x – 15
Answer: The quotient is x + 3. The remainder is 0.
Use long division to find (x2 + 5x + 6) ÷ (x + 3).
A. x + 2
B. x + 3
C. x + 2x
D. x + 8
Divide Polynomials
Which expression is equal to (a2 – 5a + 3)(2 – a)–1?
A a+3
B
C
D
Divide Polynomials
Read the Test Item
Since the second factor has an exponent of –1,
this is a division problem.
Solve the Test Item
Rewrite 2 – a as –a + 2.
–a(–a + 2) = a2 – 2a
–5a – (–2a) = –3a
3(–a + 2) = –3a + 6
Subtract. 3 – 6 = –3
Divide Polynomials
The quotient is –a + 3 and the remainder is –3.
Therefore,
Answer: The answer is D.
.
Which expression is equal to (x2 – x – 7)(x – 3)–1?
A.
B.
C.
D.
Synthetic Division
Use synthetic division to find
(x3 – 4x2 + 6x – 4) ÷ (x – 2).
Step 1 Write the terms of the
x3 – 4x2 + 6x – 4
dividend so that the degrees  


of the terms are in
6 –4
descending order. Then write 1 –4
just the coefficients as
shown.
Step 2 Write the constant r of
1 –4
6 –4
the divisor x – r to the
left. In this case, r = 2.
1
Bring the first
coefficient, 1, down as
shown.
Synthetic Division
Step 3 Multiply the first
coefficient by r : 1 ● 2 = 2.
Write the product under
the second coefficient.
Then add the product and
the second coefficient:
–4 + 2 = –2.
Step 4 Multiply the sum, –2, by
r : –2 ● 2 = –4. Write the
product under the next
coefficient and add:
6 + (–4) = 2.
1 –4
6 –4
2
1 –2
1 –4
6 –4
2
–4
1 –2
2
Synthetic Division
Step 5 Multiply the sum, 2, by
r : 2 ● 2 = 4. Write the
product under the next
coefficient and add:
–4 + 4 = 0. The remainder
is 0.
1 –4
6
2 –4
1 –2
2
The numbers along the bottom are the coefficients of
the quotient. Start with the power of x that is one less
than the degree of the dividend.
Answer: The quotient is x2 – 2x + 2.
–4
4
0
Use synthetic division to find (x2 + 8x + 7) ÷ (x + 1).
A. x + 9
B. x + 7
C. x + 8
D. x + 7
Divisor with First Coefficient Other than 1
Use synthetic division to find
(4y3 – 6y2 + 4y – 1) ÷ (2y – 1).
Rewrite the divisor so it has a leading coefficient of 1.
Divide numerator and
denominator by 2.
Simplify the
numerator and
denominator.
Divisor with First Coefficient Other than 1
The result is
.
Divisor with First Coefficient Other than 1
Answer: The solution is
Check:
.
Divide using long division.
2y2 –2y + 1
4y3 – 2y2
–4y2 + 4y
–4y2 + 2y
2y – 1
2y – 1
0
The result is
.
Use synthetic division to find
(8y3 – 12y2 + 4y + 10) ÷ (2y + 1).
A.
B.
C.
D.