Kuta Software - Infinite Algebra 2 Name___________________________________

The Rational Root Theorem Date________________ Period____

State the possible rational zeros for each function.

1) f ( x) = 3 x 2 + 2 x 1 2) f ( x) = x 6 64

3) f ( x) = x 2 + 8 x + 10 4) f ( x) = 5 x 3 2 x 2 + 20 x 8

5) f ( x) = 4 x 5 2 x 4 + 30 x 3 15 x 2 + 50 x 25 6) f ( x) = 5 x 4 + 32 x 2 21

7) f ( x) = x 3 27 8) f ( x) = 2 x 4 9 x 2 + 7

State the possible rational zeros for each function. Then find all rational zeros.

9) f ( x) = x 3 + x 2 5 x + 3 10) f ( x) = x 3 13 x 2 + 23 x 11

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 11) f ( x) = x 3 + 4 x 2 + 5 x + 2 12) f ( x) = 5 x 3 + 29 x 2 + 19 x 5

13) f ( x) = 4 x 3 9 x 2 + 6 x 1 14) f ( x) = 3 x 3 + 11 x 2 + 5 x 3

15) f ( x) = 5 x 4 46 x 3 + 84 x 2 50 x + 7 16) f ( x) = 3 x 4 10 x 3 24 x 2 6 x + 5

17) f ( x) = 3 x 3 + 9 x 2 + 4 x + 12 18) f ( x) = 2 x 3 + 9 x 2 + 19 x + 15

Critical thinking question:

19) In the process of solving 2 x 3 + 7 x 2 + 9 x + 10 = 0 you test 1, 2, 5, and 10 as possible zeros and determine
5
that none of them are actual zeros. You then discover that is a zero. You calculate the depressed
2
3
polynomial to be 2 x + 2 x + 4. Do you need to test 1, 2, 5, and 10 again? Why or why not?

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 Kuta Software - Infinite Algebra 2 Name___________________________________

The Rational Root Theorem Date________________ Period____

State the possible rational zeros for each function.

1) f ( x) = 3 x 2 + 2 x 1 2) f ( x) = x 6 64
1 1, 2, 4, 8, 16, 32, 64
1,
3

3) f ( x) = x 2 + 8 x + 10 4) f ( x) = 5 x 3 2 x 2 + 20 x 8
1, 2, 5, 10 1 2 4 8
1, 2, 4, 8, , , ,
5 5 5 5

5) f ( x) = 4 x 5 2 x 4 + 30 x 3 15 x 2 + 50 x 25 6) f ( x) = 5 x 4 + 32 x 2 21
1 5 25 1 5 25 1 3 7 21
1, 5, 25, , , , , , 1, 3, 7, 21, , , ,
2 2 2 4 4 4 5 5 5 5

7) f ( x) = x 3 27 8) f ( x) = 2 x 4 9 x 2 + 7
1, 3, 9, 27 1 7
1, 7, ,
2 2

State the possible rational zeros for each function. Then find all rational zeros.

9) f ( x) = x 3 + x 2 5 x + 3 10) f ( x) = x 3 13 x 2 + 23 x 11
Possible rational zeros: 1, 3 Possible rational zeros: 1, 11
Rational zeros: {3, 1 mult. 2} Rational zeros: {1 mult. 2, 11}

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 11) f ( x) = x 3 + 4 x 2 + 5 x + 2 12) f ( x) = 5 x 3 + 29 x 2 + 19 x 5
Possible rational zeros: 1, 2 1
Possible rational zeros: 1, 5,
Rational zeros: {1 mult. 2, 2} 5

Rational zeros: { 1
5
, 5, 1 }
13) f ( x) = 4 x 3 9 x 2 + 6 x 1 14) f ( x) = 3 x 3 + 11 x 2 + 5 x 3
1 1 1
Possible rational zeros: 1, , Possible rational zeros: 1, 3,
2 4 3

Rational zeros: 1 mult. 2, { 1

4 } Rational zeros: { 1
3
, 3, 1 }

15) f ( x) = 5 x 4 46 x 3 + 84 x 2 50 x + 7 16) f ( x) = 3 x 4 10 x 3 24 x 2 6 x + 5
1 7 1 5
Possible rational zeros: 1, 7, , Possible rational zeros: 1, 5, ,
5 5 3 3

Rational zeros: { 1
5
, 7, 1 mult. 2 } Rational zeros: { 1
3
, 5, 1 mult. 2 }

17) f ( x) = 3 x 3 + 9 x 2 + 4 x + 12 18) f ( x) = 2 x 3 + 9 x 2 + 19 x + 15
Possible rational zeros: Possible rational zeros:
1 2 4 1 3 5 15
1, 2, 3, 4, 6, 12, , , 1, 3, 5, 15, , , ,
3 3 3 2 2 2 2
Rational zeros: {3}
Rational zeros: { } 3
2

Critical thinking question:

19) In the process of solving 2 x 3 + 7 x 2 + 9 x + 10 = 0 you test 1, 2, 5, and 10 as possible zeros and determine
5
that none of them are actual zeros. You then discover that is a zero. You calculate the depressed
2
3
polynomial to be 2 x + 2 x + 4. Do you need to test 1, 2, 5, and 10 again? Why or why not?
No. That would be like factoring 740 and discovering 3 isn't a factor but then checking if
anything 740 breaks down into has a factor of 3. If the original problem doesn't have a
factor of 3 then nothing it factors into will have a factor of 3.

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p X2q0s1y2G eKFuztTay 0Seoqfrtqwuaor4er zLGLdC6.6 f 8AJlvlF DrRi9guhXtAsQ Gr3eXsueSrVvEeGdy.M 3 rMNaTdge0 8wEiZtLhi 0INnKfNisnFiJtyeb 6A4l5g4epbErIa0 W2t.R -2- Worksheet by Kuta Software LLC