Precalculus 2.3 Real Zeros of Polynomial Functions

Precalculus 2.3a

I. Long Division of Polynomials:

A. Recall: When you did long division

Example 1:

625/5 =

As you can see, 5 goes into 625 perfectly and there is no remainder so this means that 5 is a factor of 625.

Example 2:
625/4 =


As you can see, 4 does not go into 625 perfectly, there is a remainder of 1 so we would write the answer like

156 + 1/4

You write the answer or quotient plus the remainder over the divisor

B. Long Division of Polynomials:

Now try doing the same concept only using Polynomials:

Example 3:


So therefore since (x - 4) is a factor so

x - 4 = 0
x = 4

Therefore (x - 4) is a factor and since there is not a remainder, then (x - h) is a factor and x = h.

Example 4:

Therefore (x + 2) is a factor and since there is not a remainder, then (x - h) is a factor and x = h,
then x = -2

Example 5:




Since when you divide by (x + 1), it has a remainder, it is not a factor and x = -1 is not a zero.

Example 6:





So again, since (x - 2) is not a factor, x = 2 is not a zero.

Precalculus 2.3b Synthetic Division:
To divide ax³ + bx² + cx + d by (x - k), use the following pattern:

Example 1: (5x² - 17x - 12) / (x - 4)

Therefore, (5x² - 17x - 12) /(x - 4) = 5x + 3

since there is not a remainder!

Example 2: (x⁴ + 5x³ + 6x² - x - 2) / (x + 2)

Again, since the remainder is 0,

(x⁴ + 5x³ + 6x² - x - 2) / (x + 2) =

x³ + 3x² - 1

Example 3: (4x³ - 13x + 10) / (x + 1)

Since there is not an x², we put zero in that place:

Since 19 is the remainder:

(4x³ - 13 x + 10) / (x + 1) = 4x² - 4x - 9 + 19/(x + 1)

C. The Remainder Theorem: If a polynomial f (x) is divided by (x - k), the remainder is r = f (x)

Recall from Example 3, we saw the remainder = 19 so...

f (-1) = 4(-1)³ - 13 (-1) + 10= 4 (-1) + 13 + 10= -4 + 13 + 10= 19

this gives the same answer!!

Let's try another:

Example 4: (4x³ - 13x + 10) / (x - 2)

f(2) = 4(2)³ - 13(2) + 10 = 16, again, the same answer!

D. The Factor Theorem:
A polynomial f(x) has a factor (x - k) if and only if f(k) = 0.

In Summary:

The remainder "r", obtained in the synthetic division of f (x) by (x - k) provides the following information:
1. the remainder "r" gives the value of "f" at x = k. That is r = f (k)

2. If r = 0, then (x - k) is a factor of f (x)

3. If r = 0, then (k, 0) is an x-intercept of the graph of f (x).

II. The Rational Zero Test:
relates the possible rational zeros of a polynomial

Rational zero = p/q

where "p" and "q" have no common factors other than 1, "p" is a factor of the constant a₀ and q is a factor of the leading coefficient a_n
polynomial:

f(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₂x² + a₁x + a₀

Possible rational zeros = (factors of constant)/(factors of leading coefficient)

Example 1: f (x) = 4x⁵ - 8x⁴ - 5x³ + 10x² + x - 2

So the possible rational zeros = (1, -1, 2, -2) / (1, -1, 2, -2, 4, -4) so

The possible zeros are : {1, -1, 1/2, -1/2, 1/4, -1/4, 2, -2}

Looking at the graph of the function, we see that the real zeros are:
(-1, 0), (-1/2, 0), (1/2, 0), (1, 0), and (2, 0)

You can also use synthetic Division to eliminate possible zeros:
f (x) = 4x⁵ - 8x⁴ - 5x³ + 10x² + x - 2

try x = -1,

so f(x) = (x + 1)(4x⁴ - 12x³ + 7x² + 3x - 2)


so f (x) = (x - 1)(x + 1)(4x³ - 8x² - 1x + 2)


Therefore f(x) = (x - 1)(x + 1)(x -2)(4x² - 1)

f(x) = (x -1)(x + 1)(x -2)(2x + 1)(2x - 1)

so x = -1, x = 1, x = 2, x = -1/2, x = 1/2

so you can see they are the same answers.

III. Third Test for zeros:

A real Number "b" is an upper bound for the real zeros of "f" if no zeros are greater than "b". Similarily, "b" is a lower bound if no real zeros of "f" are less than "b"

Example 1: f(x) = x⁴ - 4x³ + 15

1. degree is even

2. Leading coefficient is positive

so it rises to the left and rises to the right

3. points (-3, 204), (-2, 63), (-1, 20), (0, 15), (1, 12), (2, -1), (3, -12), (4, 15)

So from the points we chose, the lower bound will be 1 and the upper bound is 4

We can use synthetic division to verify these thoughts:

Let x = -1, you can see that the remainder is 20 and the last row is:

1, -5, 5, -5, 20

Let x = 4, you can see that the remaider is 15 and the last row is:

1, 0, 0, 0, 15

Let x = 1, you can see that the remainder is 12 and the last row is:

1, -3, -3, -3, 12

Therefore we can conclude, given x = c:

1. If c is greater than zero and each number in the last row is either positive or zero, "c" is an upper bound for the real zeros of "f".

2. If c is less than zero and each number in the last row are alternately positive and negative, (zero counts as either positive or negative ), "c" is a lower bound for the real zeros of "f".

Therefore -1 is a lower bound and 4 is an upper bound.