Practice 1 Solutions

Determine if the following are polynomials. If so, classify by degree and number of terms.

  1. 3x2y+2xy–4

Yes, this is a third degree polynomial.
1st term exp:
2 + 1 = 3
2nd term: 1 + 1 = 2
3rd term: 0

Note

Q: How do you find the degree when the expression contains more than one variable?
A: Add the degree (exponents) of all of the variables.

  1. x4y–2+5x+1

No, this is not a polynomial because the first term has a negative exponent.

  1. 2x2+4x–xy

No, this is not a polynomial because xy is under a radical.

  1. x5+5x4+10x3+10x2+5x+1

Yes, this is a 5th degree polynomial.
1st term: largest exponent: 
5

Quintic polynomial with 6 terms

Simplify. Write answers in standard form.

  1. x2xy+7–x+1x–1

x3y+7x2–x2–1x3y+7x2–x2+1

x3y+6x2+1

Note

Q: What property allows you to multiply binomials together?
A: The distributive property

  1. ab+2ab–3–ab+4

a2b2–3ab+2ab–6–ab+4

a2b2–2ab–2

  1. g+5g2–3g+2

g3–3g2+2g+5g2–15g+10

g3+2g2–13g+10

  1. 2xy7x2y–2–3xx2y2+5

14x3y2–4xy–3x3y2–15x

11x3y2–4xy–15x

Note

Q: What is the degree of the simplified polynomial? Explain.
A: Degree 5, because 3+2=5.

  1. k+2k–2+k+3k–3

k2–4+k2–9

2k2–13

  1. 2r+1r2+4r–2

2r3+8r2–4r+r2+4r–2

2r3+9r2–2

Note

Q: Why is it not possible to combine terms with different degrees?
A: Because they are not like terms and only like terms can be added.

Factor completely.

Note

Problems 11–16
Q: What is the first step when factoring?
A: Determine if there is a GCF.

  1. 3x2y2–5xy–12

3xy+4xy–3

  1. 4a2b2–4a2–9b2+9

4a2b2–1–9b2–14a2–9b2–1

2a+32a–3b+1b–1

Note

Q: What special product occurs in this polynomial?
A: A difference of two squares

  1. 5m2+70m+245

5m2+14m+49

5m+72

Note

Q: What special product occurs in this polynomial?
A: A perfect square trinomial

  1. 4x2–26x+30

22x2–13x+15

22x–3x–5

  1. 35q4–8q3–3q2

q235q2–8q–3

q25q+17q–3

  1. 2xy2+8x–9y2–36

2xy2+4–9y2+4

2x–9y2+4

Factor the sum or difference of cubes completely.

Note

Problems 17–20
It may be helpful to write each term as a base to the 3rd power before using the factoring patterns for the sum and difference of cubes.

  1. 8y3–1

8y=2y3, 1=132y–12y2+2y·1+12

2y–14y2+2y+1

  1. 27c3+64

27c3=3c3, 64=433c+43c2–3c·4+42

3c+49c2–12c+16

  1. 250v3+16

2125v2+8; 125v3=5v3, 8=2325v+25v2–5v·2+22

25v+225v2–10v+4

  1. x3y3–a3b3

x3y3=xy3, a3b3=ab3xy–abxy2+xy·ab+ab2

xy–abx2y2+abxy+a2b2

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