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In Zeros of Polynomials, we saw how factoring was used to solve
polynomial expressions set equal to zero.
This section will focus on identifying the zeros of quadratic and cubic polynomial functions. We will be extending our work with the Zero Product Property, by working with factorable functions.
Warm up those factoring skills!
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A polynomial function is a function which is defined by a polynomial.
(In plain English, it's a polynomial expression set equal to P(x), or some other notation.)
Examples: P(x) = x2 + x - 6; P(x) = x3 - x2 - 12x; P(x) = x2 + 4x + 4 |
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The zeros of a polynomial function, P(x), will be all of the values of x that make the polynomial equal zero. They are the x values that satisfy P(x) = 0.
Quadratic Polynomial Functions:
A quadratic polynomial function is a function in which the highest power (exponent) is two.
For example, P(x) =x2 - 3x - 10 is a quadratic polynomial function.
Remember: (x + a)(x + b) = x2 + (a + b)x + ab
If set equal to zero, we have x2 + (a + b)x + ab = 0 and the zeros are x = -a, and x = -b.
(since the factors set equal to zero are (x + a) = 0 and (x + b) = 0)
What are the zeros of the function
P(x) = x2 - 4x + 3 ?
Solution:
Set P(x) = 0 and factor.
P(x) = x2 - 4x + 3 = 0
x2 - 4x + 3 = 0
(x - 3)(x - 1) = 0
Set factors = 0 to find zeros.
(x - 3) = 0
x = 3 (zero)
(x - 4) = 0
x = 4 (zero)
Zeros: {3, 4}
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What are the zeros of the function
P(x) = (x - 5)2 - 81 ?
Solution:
Set P(x) = 0 and factor.
Notice the difference of two squares.
P(x) = (x - 5)2 - 81 = 0
((x - 5) - 9)((x - 5) + 9) = 0
Set factors = 0 to find zeros.
(x - 5) - 9 = 0
x - 14 = 0
x = 14 (zero)
(x - 5)+ 9 = 0
x + 4 = 0
x = -4 (zero)
Zeros: {-4, 14} |
Cubic Polynomial Functions:
A cubic polynomial function is a function in which the highest power (exponent) is three.
For example, P(x) = x3 + 6x2 + 11x + 6 is a cubic polynomial function.
In Algebra 1, when working with a cubic function, the linear and quadratic factors may be given, so the P(x) shown above would be stated as P(x) = (x + 3)(x2 + 3x + 2).
You will not be asked to factor a cubic.
What are the zeros of the function
P(x) = (x + 3)(x2 + 3x + 2) ?
Solution:
Set factors = 0.
(x + 3) = 0
x = -3 (zero)
(x2 + 3x + 2) = 0
(x + 2)(x + 1) = 0
(x + 2) = 0
x = -2 (zero)
(x + 1) = 0
x = -1 (zero)
Zeros: {-3, -2, -1}
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What are the zeros of the function
P(x) = x3 - 12x2 + 32x ?
Solution:
Set P(x) = 0 and factor.
Notice the GCF.
P(x) = x3 - 12x2 + 32x = 0
x3 - 12x2 + 32x = 0
x(x2 - 12x - 32) = 0
x(x - 8)(x - 4) = 0
Set factors = 0 to find zeros.
x = 0 (zero)
(x - 8) = 0
x = 8 (zero)
(x - 4) = 0
x = 4 (zero)
Zeros: {0, 4, 8} |
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