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## Math121, Chapter 2.4 Notes

by: Mallory McClurg

69

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4

# Math121, Chapter 2.4 Notes Math 121

Marketplace > University of Mississippi > Math > Math 121 > Math121 Chapter 2 4 Notes
Mallory McClurg
OleMiss
GPA 3.37

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Here are examples of every type of higher degree polynomial equations we need to know how to solve, with step-by-step explanations of every single one! You may need this to study for the quiz.
COURSE
College Algebra
PROF.
Dirle
TYPE
Class Notes
PAGES
4
WORDS
CONCEPTS
polynomial
KARMA
25 ?

## Popular in Math

This 4 page Class Notes was uploaded by Mallory McClurg on Wednesday September 7, 2016. The Class Notes belongs to Math 121 at University of Mississippi taught by Dirle in Fall 2016. Since its upload, it has received 69 views. For similar materials see College Algebra in Math at University of Mississippi.

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Date Created: 09/07/16
Math121 Chapter 2 Notes Lesson 2.4 – Higher Degree Polynomial Equations Example 1. 3 2 5x + 5x = 10x (First, subtract 10x from both sides.) 5x + 5x – 10x = 0 (Now, we can factor out an “x” from each term on the left side of the20.) x (5x + 5x -10) = 0 (This basically gives us a quadratic equation that we can factor out like normal. Just, this time, “x” will equal three different things, instead of just 2 like it always was in the last lesson.) x (5x – 5) (x + 2) (The “x” out by itself in the front is equal to zero. We can set the two factored sets equal to zero, also, to solve for “x”.) x = 0 x = 1 x = -2 Example 2. x + 5x + 16x – 80 = 0 (When you have a constant at the end of this kind of equation, you must look at this in pairs. x and 5x are the first pair, and 16x and -80 are the second pair. What common factors do you see within the pairs? Bring 2 them outside the parentheses.) x (x + 5) and 16 (x – 5) (At this point, we’re going to look back at the pairs, and divide each of them by their respective common factors.) 3 2 2 (x + 5x ) / x and (16x – 80) / 16 (And when you cross out the numbers and variables that are similar in the numerator and denominator of each, both equations simplify to…. “x + 5”. This means, in this equation, we know that (x + 5) is going to be one of the expressions when factoring out our equation to solve for x. The other expression is going to be the difference of the two common factors we found in the second 2 step. So “x – 16” is going to give us our other two expressions.) (x – 16) (x + 5) (We should recognize that “x – 16” is the difference of two squares, like we learned in the last lesson. That means this factors out to “x – 4” and “x + 4”. So we have our 3 expressions… (x + 5), (x – 4) and (x + 4). Set each equal to zero and solve for x.) x = -5 x = 4 x = -4 Example 3. 2 (x – 4) – 7 (x – 4) + 6 = 0 (First, FOIL out the “(x – 4) ”. Then distribute the -7 to the x and the -4. Write it all out if you have to!) 2 x – 4x – 4x + 16 – 7x + 28 + 6 = 0 (Combine like terms.) x – 15x + 50 = 0 (Now, factor it out and then solve each expression for zero.) (x – 5) (x – 10) = 0 x = 5 x = 10 Example 4. (y – 1) – 13 (y – 1) + 30 = 0 (First, substitute the expression in the parentheses with the variable A.) A – 13A + 30 = 0 (Does that form look familiar? Factor it out like a quadratic equation.) (A – 3) (A – 10) = 0 (Solve for A.) A = 3 A = 10 (So, remember we used “A” to substitute for “y – 1”. Now, we substitute it back in.) 2 2 y – 1 = 3 y – 1 = 10 (Add “1” to both sides of both equations.) y = 4 y = 11 (Take the square root of each side.) y = +/- 2 y = +/- √11 (These are our 4 answers!) Ex2mple 5.2 2 (x – 6x) + 2 (x – 6x) – 15 = 0 (Substitute “A” for the expression in the parentheses.) A + 2A – 15 = 0 (Factor this out like a quadratic equation.) (A – 3) (A + 5) = 0 (Set each equal to zero and solve for A.) A = 3 A = -5 (Substitute “x – 6x” back in for A.) x – 6x = 3 x – 6x = -5 (Put each into its proper ax +bx+c=0 form.) 2 2 x – 6x – 3 = 0 x – 6x + 5 = 0 (Try to factor out both equations. The first doesn’t factor, so we have to use the quadratic formula for that one. The second one can be factored easily.) 2 x = (6 +/- √(-6) – 4(1)(-3)) / 2(1) (x – 5) (x – 1) = 0 x = (6 + √48) / 2 x = (6 - √48) / 2 x = 5 x = 1 (The square roots can be simplified, because √48 = √2*2*2*2*3 = (2*2)√3 = 4√3.) x = (6 + 4√3) / 2 x = (6 – 4√3) / 2 x = 5 x = 1 (We can simplify the fractions in the first two solutions by factoring a 2 out of the numerator and cancelling it out in the denominator.) x = 3 + 2√3 x = 3 - 2√3 x = 5 x = 1 Example 6. 1/2 1/4 z – 7z + 12 = 0 (The first two terms in this equation have a common factor of “z ”, so rewrite the problem with the common factors in parentheses.) (z ) – 7(z ) + 12 = 0 (You should be able to recognize that this now looks like 2 the quadratic A – 7A + 12 = 0 equation, which can be factored out.) (A – 4) (A – 3) = 0 1/4 A = 4 A = 3 (Substitute “z ” back in for A.) z1/4= 4 z1/4= 3 (Raise both sides of both equations to an exponent of 4, to get rid of “1/4” the ex4onent 4 .) z = 4 = 256 z = 3 = 81 Example 7. x7/2 – 5x 5/2+ 6x 3/2= 0 (What common factor do you see between the three values in th3/2quation? You can take out an “x ”.) x3/2(x – 5x + 6) = 0 (Factor out the quadratic equation, and set your three expressions to zero to solve for x.) x3/2(x – 3) (x – 2) = 0 x = 0 x = 3 x = 2

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