Practice 2 Solutions

Simplify. State when the given expression is undefined.

  1. 3xx2+7x+10·x+2x2+5x

3xx+5x+2·x+2xx+53xx+5x+2·x+2xx+5

3x+52, x≠–5, –2, 0

  1. b2–9b+143b+18÷3b2–15b–726b+36·b2+6b+9b2–b–2

b–7b–23b+6÷3b2–5b–246b+6·b+3b+3b–2b+1b–7b–23b+6·6b+63b–8b+3·b+3b+3b–2b+1b–7b–23b+6·62b+63b–8b+3·b+3b+3b–2b+1

2b–7b+33b–8b+1, b≠–6, –3, –1, 2, 8 

  1. 2n–3n3+4n2+n+4÷4n2–94n2+22n+24

2n–3n2n+4+1n+4÷2n+32n–322n2+11n+122n–3n2+1n+4·22n+3n+42n+32n–32n–3n2+1n+4·22n+3n+42n+32n–3

2n2+1, n≠–4, ±32

Note

For the expression n2+1 there are no restrictions in the set of real numbers. You will learn more about the square root of negative numbers in later lessons.

  1. k2–j2k–j÷k4–j4k3–j3

k+jk–jk–j÷k2+j2k2–j2k–jk2+kj+j2k+jk–jk–j·k–jk2+kj+j2k2+j2k+jk–jk+jk–jk–j·k–jk2+kj+j2k2+j2k+jk–j

k2+kj+j2k2+j2, k≠±j

Note

k2+j2 cannot equal zero in the real number system.

Simplify.

  1. x2+4x–5x2–3x–18x2+6x+5x2–4x–12

x2+4x–5x2–3x–18÷x2+6x+5x2–4x–12x+5x–1x–6x+3÷x+5x+1x–6x+2x+5x–1x–6x+3·x–6x+2x+5x+1x+5x–1x–6x+3·x–6x+2x+5x+1x≠–5, –3, –2, –1, 6

x–1x+2x+3x+1, x≠–5, –3, –2, –1, 6 

  1. 3x2+x–142x2+13x+20·3x2–4810x2–24x+8÷18x2–9810x2+21x–10

3x+7x–22x+5x+4·3x2–1625x2–12x+4÷29x2–492x+55x–23x+7x–22x+5x+4·3x+4x–425x–2x–2÷23x+73x–72x+55x–23x+7x–22x+5x+4·3x+4x–425x–2x–2·2x+55x–223x+73x–7

3x+7x–22x+5x+4·3x+4x–425x–2x–2·2x+55x–223x+73x–7x≠–4, –52, ±73, 25, 2 

3x–443x–7, x≠–4, –52, ±73, 25, 2 

  1. 8h3–16h2+6h·6h4+12h3+6h24h2–1

2h–14h2+2h+16hh+1·6h2h2+2h+12h+12h–12h–14h2+2h+16hh+1·6h2h+1h+12h+12h–12h–14h2+2h+16hh+1· 6h2 h+1h+12h+12h–1h≠–1, 0, ±12

hh+14h2+2h+12h+1, h≠–1, 0, ±12 

  1. r2+3r–10r2+6r+9÷r2+r–2020r2–11r–3÷5r2–9r–2r2–r–12

r+5r–2r+3r+3÷r+5r–44r–35r+1÷5r+1r–2r+3r–4r+5r–2r+3r+3·4r–35r+1r+5r–4·r+3r–45r+1r–2r+5r–2r+3r+3·4r–35r+1r+5r–4·r+3r–45r+1r–2r≠–5, –3, –15, 34, 4

4r–3r+3, r≠–5, –3, –15, 34, 4 

  1. 4w2w2+8w+16÷165w2–80

4w2w+4w+4÷165w2–164w2w+4w+4÷165w+4w–44w2w+4w+4·5w+4w–4164w2w+4w+4·5w+4w–4164 w≠±4

5w2w–44w+4, w≠±4 

  1. y2–5y+6y2+3y+2y2–2y–3y2–4

y2–5y+6y2+3y+2÷y2–2y–3y2–4y–2y–3y+1y+2÷y–3y+1y+2y–2y–2y–3y+1y+2·y+2y–2y–3y+1y–2y–3y+1y+2·y+2y–2y–3y+1y≠±2, –1, 3

y–22y–12, y≠±2, –1, 3

  1. Find the length of a rectangle when the width is x2+2x+1 x2+8x+15 meters and the area is x2–1 x2+10x+25   meters2 . Explain whether x = 1 is a possible solution.

A=lwA÷w=ll=x2–1x2+10x+25÷x2+2x+1x2+8x+15l=x+1x–1x+5x+5÷x+1x+1x+5x+3l=x+1x–1x+5x+5·x+5x+3x+1x+1l=x+1x–1x+5x+5·x+5x+3x+1x+1x≠–5, –3, –1

l=x–1x+3x+5x+1meters

Yes, x = 1 because it is not a restriction (or excluded value).

  1. Find the area of a triangle with base 2x2+5x+3 x–3 feet and the height of x2–x–6 x2–1 feet. Then determine if a length of x = 3 is possible. Explain your reasoning.

A=12bhA=12·2x2+5x+3x–3·x2–x–6x2–1A=12·2x+3x+1x–3·x–3x+2x+1x–1A=12·2x+3x+1x–3·x–3x+2x+1x–1x≠±1, 3

A=2x+3x+22x–1 ft2

It is not possible for x = 3 because it would cause the denominator of the area to equal zero.

  1. When rational expressions are divided, the    reciprocal     of the term    after     the division symbol must be written.
  1. When a rational expression is    undefined  , the denominator has a value of zero.

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