# Linear Factors

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```Name: ________________________________________
Period: ______ Date: _______
WS • 5-1 Polynomials, Linear Factors & Zeros
Write each polynomial in standard form. Then classify it by degree and by number of terms.
1. x2 + 3x  4x3
2. a3(a2 + a + 1)
3. x(x + 5)  5(x + 5)
4. 6  2x3  4 + x3
Determine the end behavior of the graph of each polynomial function.
5. y = 3x4 + 6x3  x2 + 12
6. y = 50  3x3 + 5x2 7. y = 12x4  x + 3x7  1
8. y = 20  5x6 + 3x  11x3
Determine the sign of the leading coefficient and the degree of the polynomial function for each graph.
9.
10.
11.
Write each polynomial in factored form. Check by multiplication.
12. 2x3 + 10x2 + 12x
13. x4  x3  6x2
14. 3x3 + 18x2  27x
Find the zeros of each function. Then graph the function.
16. y = (x + 1)(x  1)(x  3)
17. y = (x + 4)2(x + 1)
Write a polynomial function in standard form with the given zeros.
19. x = 1, 3, 4
20. x = 1, 1, 2
21. x = 3, 0, 0, 5
15. x3  2x2 + x
18. y = x(x  2)(x + 5)
22. x = 1, 5, 2
23. Find the zeros of each polynomial below. For each corresponding row, shade in each number that is a
zero. The illustration made from shading the squares suggests the answer to the riddle below.
A. P(x) = x(x2  1)
B. P(x) = x(x + 2)(x + 1)(x2 + 2x  3)
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___________________________
C. P(x) = x(x + 4)(x + 3)(x + 1)(x  1)
D. P(x) = x(x2  25)(x2 + 4x + 3)
___________________________
____________________________
E. P(x) = (x2 + x  20)(x + 2)(x2 + 4x + 3)
F . P(x) = (x2  9)(x2  25)
_____________________________
_____________________________
G. P(x) = (x2 + 9x + 20)(x2  5x + 6)(x  5)
H. P(x) = (x2  5x + 6)(x2  9x + 20)
______________________________
______________________________
I. P(x) = x2  6x + 9
J. P(x) = (x2  4x + 4)(x2  4x + 4)
________________________________
_______________________________
K. P(x) = x(x  2x + 1)(x  2)
2
____________________________
A
–5
–4
–3
–2
–1
0
1
2
3
4
5
B
5
–5
–4
–3
–2
–1
0
1
2
3
4
C
4
5
–5
–4
–3
–2
–1
0
1
2
3
D
3
4
5
–5
–4
–3
–2
–1
0
1
2
E
2
3
4
5
–5
–4
–3
–2
–1
0
1
F
1
2
3
4
5
–5
–4
–3
–2
–1
0
G
0
1
2
3
4
5
–5
–4
–3
–2
–1
H
–1
0
1
2
3
4
5
–5
–4
–3
–2
I
–2
–1
0
1
2
3
4
5
–5
–4
–3
J
–3
–2
–1
0
1
2
3
4
5
–5
–4
K
–4
–3
–2
–1
0
1
2
3
4
5
–5
Riddle: This grows above the ground, but the solutions to the polynomials above lie beneath. And as it grows, it
provides shade to those underneath. What is it?
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