College Algebra

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Open Chapter Ch. 1: Graphs, Functions, and Models
Lesson #1 Graphing
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Lesson #2 Graphing Functions
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Lesson #3 Slope of Linear Functions
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Lesson #4 Linear Equations and Modeling
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Lesson #5 Linear Equations, Functions, Zeros and Applications
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Lesson #6 Linear Inequalities
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Open Chapter Ch. 2: More on Functions
Lesson #7 Analyzing Functions
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Exam Exam 1
Lesson #8 Algebra for Functions
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Lesson #9 The Composition of Functions
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Lesson #10 Symmetry
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Lesson #11 Transformations
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Lesson #12 Equations of Variation
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Open Chapter Ch. 3: Quadratic Functions and Equations; Inequalities
Lesson #13 Complex Numbers
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Lesson #14 Quadratic Equations, Functions, Zeros and Models
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Lesson #15 Analyzing Graphs of Quadratic Functions
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Lesson #16 Rational and Radical Equations
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Exam Midterm Exam
Lesson #17 Absolute Value Equations and Inequalities
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Open Chapter Ch. 4: Polynomial Functions and Rational Functions
Lesson #18 Polynomial Functions and Models
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Lesson #19 Graphing Polynomial Functions
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Lesson #20 Polynomial Division
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Lesson #21 The Zeros of Polynomial Functions
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Lesson #22 Rational Functions
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Lesson #23 Polynomial and Rational Inequalities
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Open Chapter Ch. 5: Exponential Functions and Logarithmic Functions
Lesson #24 Inverse Functions
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Lesson #25 Exponential Functions and Graphs
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Lesson #26 Logarithmic Functions and Graphs
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Exam Exam 3
Lesson #27 Properties of Logarithmic Functions
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Lesson #28 Solving Exponential Equations and Logarithmic Equations
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Lesson #29 Applications and Models: Growth and Decay; Compound Interest
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Open Chapter Ch. 6: Systems of Equations
Lesson #30 Systems of Equations in Two Variables
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Lesson #31 Systems of Inequalities and Linear Programming
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Lesson #32 Nonlinear Systems of Equations and Inequalities
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Lesson #33 Sequences and Series
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Exam Final Exam

Assignments:

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Lesson Objectives:

- Factor Theorem
- Long and Synthentic Division
- Remainder Theorem


The Factor Theorem.

If you divide two polynomials and get 0 for the remainder, then the divisor is a factor of the dividend.

Remember that the dividend is the first value when you're dividing two numbers; or it's the number on the top of the fraction; or when you're doing long division, it's the value that's inside of the box. The divisor's on the outside, and the quotient's on top, and the remainder is what's left.

`"Dividend "= ("Divisor")("Quotient")+"Remainder"`



Let's start by reviewing long division.

Given the function `g(x) = x^3-2x^2-26x+12`, determine whether x-6 is a factor of g(x).

The Factor Theorem states that if we divide g(x) by x-6, and get a remainder of zero, then we know that x-6 is a factor of g(x).

So `x^3-2x^2-26x+12`. We have to divide this by x-6.

So start by seeing what we have to multiply x by to get `x^3`. We have to multiply it by `x^2`. And now multiply `x^2` times x-6. It's the same as when we're distributing within parentheses. So then we get `x^3-6x^2`, and we're going to subtract all of this from the `x^3-2x^2`. So `x^3-x^3 = 0`, and `-2x^2+6x^2` gives us `+4x^2`.

Then, bring down the -26x. So -26x, we have `4x^2-26x`. So what do we need to multiply x by to get `4x^2`? How about 4x? So now we need to multiply 4x by x-6. It's the same as distributing `4x(x-6)`, so that we get `4x^2-24x`. And then we have to subtract all of this, and we get `0-2x`.

And now carry down the positive 12, so we get `+12` over here. And now, what do we have to multiply x by to get -2x? How about -2? So now we need to multiply -2 to x-6. This will give us `-2x+12`, and we'll get a remainder of zero.

This is the remainder.

Now, determine whether x+4 is a factor of g(x). We're going to divide g(x) by x+4, and if we get a remainder of zero, that means that x+4 is a factor of g(x) by the Factor Theorem.

So we're going to write `x^3-2x^2-26x+12`, and divide this using long division by x+4.

Now, first, what do we have to multiply x by to get `x^3`? It's `x^2`. So now multiply `x^2` by x+4, and write it down over here. And we get `x^3+4x^2`, and then don't forget to subtract everything. So, `x^3-x^3 = 0` and `-2x^2-4x^2 = -6x^2`.

And then carry down the -26x. Now what do we multiply x by to get `-6x^2`? How about -6x? That means that we have `-6x^2-24x`, and then don't forget to subtract everything. So, we get `-2x+12`, if you carry the 12 down. And then, what do we multiply x by to get -2x? It's -2. So, `-2(x+4)` gives us `-2x-8`, and then if we subtract everything, `-2x+2x` gives us 0, and `12+8 = 20`. So, our remainder is 20, it's not zero. So x+4 is not a factor of g(x).



Synthetic division is another way to divide polynomials. Find the quotient and remainder of `(x^3-5x^2-10)/(x+5)`.

The numerator is called the dividend, and the denominator is known as the divisor. The divisor needs to be written in `x-c` form. So in other words, `x-(-5)`. Because we're going to put c in the corner, and then we're going to write down all of our coefficients of the dividend in order of decreasing powers. So that we have 1, -5, 0, and -10. Notice the zero. It's a place holder for the x term, which isn't there.

Now, start by bringing down the 1, and then multiply `(1)(-5)`, and then add straight down, so `(1)(-5) = -5` and `-5+-5 = -10`. Now multiply -10 by -5, and we'll get positive 50, and then add straight down. `0+50 = 50`. Now multiply `50*-5`, we get -250, and then add straight down, and get -260, and this is the remainder.

When the degree of the divisor is one, the degree of the quotient is one less than the degree of the dividend. So the quotient is `x^2-10x+50`, and the remainder is -260.



The Remainder Theorem.

If `x = c`, then f(c) is equal to the remainder, when `(f(x))/(x-c)`.

Remember that f(x) is the dividend, and x-c is the divisor.

The dividend is equal to the divisor times the quotient, plus the remainder. And `f(c) = (c-c)Q(x)+r`, and `c-c = 0`, so it cancels out this whole term, and we're left with `f(c) = r`.



Use synthetic division and the Remainder Theorem to find the function values.

Check your answer by plugging in.

So given the function, `f(x) = x^4-5x^3+2x+9`, find f(-1), f(4), and f(-5).

According to the remainder theorem, f(-1) is equal to the remainder if we divide this function by `x-(-1)`. We can do this using synthetic division, so put your -1 in the corner, and then write down the coefficients of your dividend in order of decreasing powers of x. So we have 1, -5, 0, as a placeholder for the x-squared term, 2, and 9. Now bring down the 1, and then multiply `1*-1` to get -1, and then add `-5+(-1)` to get -6. And then multiply `-6*-1` to get positive 6, and then add `0+6` to get 6. And then multiply `6*-1` to get -6, and then add `2+(-6)` to get -4. And then multiply `-4*-1` to get 4, and then add `9+4` to get 13. So the remainder is equal to 13, and by the remainder theorem, `f(-1) = 13`. We can check this by plugging in, so plug in `f(-1)`, and we get `(-1)^4-5(-1)^3+2(-1)+9`, and this simplifies to get 13.

Now by the remainder theorem, f(4) is equal to the remainder if we divide f(x) by `x-4`. So we can do this using synthetic division. Put 4 in the corner, and then put down the same coefficients that you used in the first part of our example. 1, -5, 0, 2, and 9. Now bring down the 1, and then multiply `1*4`, and you get 4, and `-5+4=-1`. And `-1*4=-4`, and `0-4=-4`, and `-4*4=-16`, and `-16+2=-14`. And `-14*4=-56`, and `9-56=-47`. And so, since the remainder is -47, `f(4) = -47`, by the remainder theorem. And we can check this by plugging back into our function, `f(4) = (4)^4-5(4)^3+2(4)+9`, and this does simplify to -47.



Now let's find f(-5).

By the remainder theorem, f(-5) is equal to the remainder of `(f(x))/(x-(-5))`.

So put -5 in the corner, and then put down your coefficients: 1, -5, 0, 2, and 9. And then bring down 1, multiply `1*-5`, add `-5+(-5)`. Multiply `-10*-5`, and get 50. Add `0+50` to get 50. Multiply `50*-5` to get -250. So we get -248, which we multiply by -5; this gives us 1240, which we add to 9 to get 1249. So `f(-5) = 1249`, and we can check that by plugging -5 back into our function, and we get `(-5)^4-5(-5)^3+2(-5)+9`, which does equal 1249.



Zeros and Factors of Polynomial Functions.

In this section, we'll be using synthetic division to find zeros and factors of polynomial functions.



If `f(c) = 0`, then `x = c` is a zero of the function, f(x).

And by the Remainder Theorem, `f(c) = 0` when the remainder is zero for `(f(x))/(x-c)`.



Use synthetic division to determine whether the given values are zeros of the polynomial function.

-1 and 2, given `f(x) = 9x^3+5x^2-50x+8`. -1 is a zero of our function, if `f(-1) = 0`. And by the Remainder Theorem, `f(-1) = 0`, if `(f(x))/(x-(-1))`, has a remainder of 0.

So, let's use synthetic division. We have 9, 5, -50, and 8. Notice that we didn't any placeholders this time. So bring down the 9, and then multiply `9*-1`; we get -9. Then add `5+(-9)`, and we get -4. And then multiply `-4*-1` to get 4. And then add `-50+4` to get -46. And then multiply `-46*-1` to get positive 46. And then add `46+8` to get 54. So that means that the remainder is equal to 54, and by the remainder theorem, `f(-1) = 54`. Since it doesn't equal zero, that means that -1 is not a zero of f.

Now to check if 2 is a zero, we need to see if `f(2) = 0`. And `f(2) = 0`, if `(f(x))/(x-(2))` has a remainder of zero, by the Remainder Theorem. So let's use synthetic division, and write down the same coefficients as before. Bring down the 9, multiply `9*2`, get 18, then add `18+5`, which gives you 23. And multiply `23*2`, which gives you 46, and add `-50+46`, which gives you -4. And `-4*2`, which gives you -8. And `8+(-8)=0`. So the remainder is zero, which means by the Remainder Theorem, `f(2) = 0`. So that means that 2 is a zero of f.



Use synthetic division to determine whether the given values are zeros of the polynomial function.

We're given the function `f(x) = x^3-5x^2+9x-45`. And they want us to check if i and -3i are zeros of the polynomial function. If `f(i) = 0`, then i is a zero of f. And f(i) is equal to the remainder of `(f(x))/(x-i)`.

So we have 1, -5, 9, and -45. Now bring down the 1, and then multiply `1*i` to get i, `-5+i` is `-5+i`. `(-5+i)i` and we get `-5i+i^2`. Now `i^2` is the same as -1; we have `8-5i`. Now multiply that by i, and we get `8i-5i^2`, and since `i^2` is the same as -1, this last term is the same as positive 5. `-45+8i+5 = -40+8i`. So the remainder is `-40+8i`.

And by the Remainder Theorem, we have that `f(i) = -40+8i`. It is not equal to zero. So that means that i is not a zero of f.

Now let's check if `f(-3i) = 0`. Because that would mean that -3i is a zero of f. And by the Remainder Theorem, `f(-3i)` is equal to the remainder if we divide `(f(x))/(x-(-3i))`.

So we can accomplish this through synthetic division. So bring down the 1, and multiply `1*-3i`. Then add `-5+(-3i)` Now multiply `(-5-3i)(-3i)`, and we get `15i+9i^2`, and `i^2` is the same as -1, so `-1*9=-9`. Now `9+15i-9` is just 15i. So multiply `15i*-3i` and we get `-45i^2`. Now `i^2` is the same as -1, so this is the same as positive 45. So our remainder is zero, and by the Remainder Theorem, we have `f(-3i) = 0`.

That means that -3i is a zero of f.



Now we can rewrite the Factor Theorem. For a polynomial, f(x), if `f(c) = 0`, then (x-c) is a factor of f(x).



Look for linear factors of the form (x-c), and use synthetic division to factor the polynomial function f(x). Then find the zeros of the function.

`f(x) = x^3-5x^2+3x+9`. So let's try `(x-(-1))` or `x+1`, and see if that's a factor.

Bring down the 1 and multiply `1*-1` to get -1. And `-5+(-1)=-6`, and `-6*-1=6`. And then `6+3=9`, and `9*-1=-9` and `9+-9=0`. So the remainder is equal to zero, and by the Remainder Theorem, `f(-1) = 0`. So by the Factor Theorem, `x+1` is a factor. The quotient, `x^2-6x+9` is also a factor. So we can break up f(x) down into `(x+1)(x^2-6x+9)`, and we can easily factor the trinomial. We'll get x and x to make `x^2`, and -3 and -3 multiply to make positive 9, but add to make -6x.

And now we need to solve `f(x) = 0` to find the zeros. So we have `(x+1)(x-3)(x-3) = 0`. And using the Principle of Zero Products, we have `x+1 = 0` or `x-3 = 0`. And we get `x = -1" or "3`.

So the zeros are -1 and 3.