= 1(2x)6 + 6(2x)5(–1)1 + 15(2x)4(–1)2 + 20(2x)3(–1)3 + 15(2x)2(–1)4 + 6(2x)1(–1)5 +1(–1)6
= 64x6 + 6(32x5)(–1) + 15(16x4)(1) + 20(8x3)(–1) + 15(4x2)(1) + 6(2x)(–1) + 1(1)
= 64x6 – 192x5 + 240x4 – 160x3 + 60x2 – 12x + 1
Notice how, in the first step, the first term has decreasing powers of the exponent, and the second term has increasing powers. The last step has alternating signs. This method may seem rather complicated, but it still beats multiplying it out the long way.
25. (x + 1)4 =
Solve It
26. (2y – 1)4 =
Solve It
27. (z – 1)5 =
Solve It
28. (3z + 2)5 =
Solve It
Answers to Problems on Multiplying Expressions
This section provides the answers (in bold) to the practice problems in this chapter.
1. Distribute . The answer is .
2. Distribute . The answer is .
3. Distribute . The answer is .
4. Distribute . The answer is .
5. Use FOIL to multiply . The answer is .
6. Use FOIL to multiply . The answer is .
7. Use FOIL to multiply . The answer is .
8. Use FOIL to multiply . The answer is .
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Chapter 8
Dividing the Long Way to Simplify Algebraic Expressions
In This Chapter
Dividing one or more terms by a single term
Working with binomial and trinomial divisors
Simplifying the process with synthetic division
Using long division to simplify algebraic expressions with variables and constants has many similarities to performing long division with just numbers. The variables do add an interesting twist (besides making everything look like alphabet soup) — with the exponents and different letters to consider. But the division problem is still made up of a divisor, dividend, and quotient (what divides in, what’s divided into, and the answer). And one difference between traditional long division and algebraic division is that, in algebra, you usually write the remainders as algebraic fractions.
Dividing by a Monomial
Dividing an expression by a monomial (one term) can go one of two ways.
Every term in the expression is evenly divisible by the divisor.
One or more terms in the expression don’t divide evenly.
If a fraction divides evenly — if every term can be divided by the divisor — the denominator and numerator have a common factor. For instance, in the first example in this section, the denominator, 6y, divides every term in the numerator. To emphasize the common factor business, I first factor the numerator by dividing out the 6y, and then I reduce the fraction.
As nice as it would be if algebraic expressions always divided evenly, that isn’t always the case. Often, you have one or more terms in the expression — in the fraction’s numerator — that don’t contain all the factors in the divisor (denominator). When this happens, the best strategy is to break up the problem into as many fractions as there are terms in the numerator. In the end, though, the method you use is pretty much dictated by what you want to do with the expression when you’re done.
Q.
A. y(4 – 3y + 5y2)
The numerator contains a factor matching the denominator.
Q.
A.
The last term doesn’t have a factor of 4x, so you break up the numerator into separate fractions for the division.
1.
Solve It
2.
Solve It
3.
Solve It
4.
Solve It
Dividing by a Binomial
Dividing by a binomial (two terms) in algebra means that those two terms, as a unit or grouping, have to divide into another expression. After dividing, if you find that the division doesn’t have a remainder, then you know that the divisor was actually a factor of the original expression. When dividing a binomial into another expression, you always work toward getting rid of the lead term (the first term — the one with the highest power) in the original polynomial and then the new terms in the division process. See the following example for a clearer picture of this concept.
This example shows a dividend that starts with a third-degree term and is followed by terms in decreasing powers (second degree, first degree, and zero degree, which is a constant — just a number with no variable). If your dividend is missing any powers that are lower than the lead term, you need to fill in the spaces with zeroes to keep your division lined up.
Also, if you have a remainder, remember to write that remainder as the numerator of a fraction with the divisor in the denominator.
Q.
To divide by a binomial, follow these steps:
1. Eliminate the lead term.
To do this, you determine what value you must multiply the first term in the divisor by to match the lead term. If the first term in the divisor is x and the lead term is x3, as it is in this example, you must multiply by x2, so write x2 over the lead term in the quotient and then multiply each term in the divisor by that value.
2. Subtract the two values by multiplying (in algebra, subtract means to change the signs and add; see Chapter 2 for more information).
3. Bring down the rest of the terms in the dividend. Notice that, at this point, you have a new lead term.
4. Eliminate the new lead term, as explained in Step 1.
5. Subtract again and bring down the other terms.
6. Repeat Steps 1 through 5 until you run out of terms.
If the problem divides evenly (that is, it doesn’t have a remainder), you know that the divisor is a factor of the dividend.
A. x2 – 5x + 7
1. Because the lead term is x3,youmultiply by x2 to match lead term.
2. Subtract the two expressions.
3. Bring down the rest of the terms in the dividend.
Notice that your new lead term is –5x2.
4. Multiply by –5x to match the new lead term.
5. Subtract again and bring down the other terms.
6. Repeat Steps 1 through 5 until you run out of terms.
Because there’s no remainder, you know that x – 4 was a factor of the dividend.
5.
Solve It
6.
Solve It
7.
Solve It
8.
Solve It
Dividing by Polynomials with More Terms
Even though dividing by monomials or binomials is the most commonly found division task in algebra, you may run across the occasional opportunity to divide by a polynomial with three or more terms.
The process isn’t really much different from that used to divide binomials. You just have to keep everything lined up correctly and fill in the blanks if you find missing powers. Any remainder is written as a fraction.
Q. (9x6 – 4x5 + 3x2 – 1) ÷ (x2 – 2x + 1) =
The fourth, third, and first powers are missing, so you put in zeros. Don’t forget to distribute the negative sign when subtracting each product.
A.
Here’s what the division looks like:
9. (x4 – 2x3 + x2 – 7x – 2) ÷ (x2 + 3x – 1) =
Solve It
10. (x6 + 6x4 – 4x2 + 21) ÷ (x4 – x2 + 3) =
Solve It
Simplifying Division Synthetically
Dividing polynomials by binomials is a very common procedure in algebra. The division process allows you to determine factors of an expression and roots of an equation. A quic
k, easy way of dividing a polynomial by a binomial of the form x + a or x – a is called synthetic division. Notice that these two binomials each have a coefficient of 1, the variable to the first degree, and a number being added or subtracted. Examples are x + 4 or x – 7.
To perform synthetic division, you use just the coefficients (the numbers multiplying the variables, x’s) of the polynomial’s terms being divided and the opposite of the number in the binomial.
Q. (x4 – 3x3 + x – 4) ÷ (x + 1) =
1. Place the opposite of the number in the divisor (–1 in this case) in front of the problem, in a little offset that looks like . Then write the coefficients in order, using zeroes to hold the place(s) of any powers that are missing.
2. Bring down the first coefficient (put it below the horizontal line) and then multiply it times the number in front. Write the result under the second coefficient and add the two numbers; put the result on the bottom.
3. Take this new result (the –4 in this problem) and multiply it times the number in front; then add the answer to the next coefficient.
4. Repeat this multiply-add process all the way down the line.
5. The result on the bottom of your work is the list of coefficients in the answer — plus the remainder, if you have one.
A.
1. The opposite of +1 goes in front. The coefficients are written in order.
2. Bring down the first coefficient; multiply it times the number in front. Write the result under the second coefficient, and add.
3. Take the –4 and multiply it by number in front; add the answer to the next coefficient.
4. Repeat until finished.
The answer uses the coefficients, in order, starting with one degree less than the polynomial that was divided. The last number is the remainder, and it goes over the divisor in a fraction. So to write the answer, the first 1 below the line corresponds to a 1 in front of x3, then a –4 in front of x2, and so on. The last –1 is the remainder, which is written in the numerator over the divisor, x + 1.
11. (x4 – 2x3 – 4x2 + x + 6) ÷ (x – 3) =
Solve It
12. (2x4 + x3 – 7x2 + 5) ÷ (x + 2) =
Solve It
Answers to Problems on Division
This section provides the answers (in bold) to the practice problems in this chapter.
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Don’t forget to use 0 as a placeholder when one of the powers is missing.
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Use the following breakdown to solve the problem:
Write the remainder as a fraction with the divisor in the denominator of the fraction.
Chapter 9
Figuring on Factoring
In This Chapter
Writing prime factorizations of numbers
Determining the greatest common factor (GCF)
Using factors to reduce algebraic fractions
Factoring an expression amounts to changing the form from a bunch of addition and subtraction to a simpler expression that uses multiplication and division. The change from an unfactored form to a factored form creates a single term — all tied together by the multiplication and division — that you can use when performing other processes. Here are some of the other tasks in algebra that require a factored form: reducing or simplifying fractions (see Chapter 3), solving equations (see Chapters 12–15), solving inequalities (see Chapter 16), and graphing functions (see Chapter 21).
Pouring Over Prime Factorizations
A prime factorization of a number is a listing of all the prime numbers whose product is that number. Of course, you first have to recognize which numbers are the prime numbers. The first 25 prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, and 97. Don’t worry about memorizing them. Infinitely many more primes exist, but these smaller numbers (and, actually, usually just the first ten) are the most commonly used ones when doing prime factorizations.
To write the prime factorization of a number, start by writing that number as the product of two numbers and then writing each of those two numbers as products, and so on, until you have only prime numbers in the product.
By convention, prime factorizations are written with the prime factors going from the smallest to the largest. The specific order helps when you’re trying to find common factors in two or three (or more) different numbers.
Q. Find the prime factorization of 360.
A. 360 = 23 × 32 × 5
360 = 10 × 36 = 2 × 5 × 6 × 6 = 2 × 5 × 2 × 3 × 2 × 3 = 23 × 32 × 5
Q. Find the prime factorization of 90.
A. 90 = 2 × 32 × 5
90 = 9 × 10 = 3 × 3 × 2 × 5 = 2 × 32 × 5
You can start the multiplication in different ways. For example, maybe you started writing 90 as the product of 6 and 15. How you start doesn’t matter. You’ll always end up with the same answer.
1. Write the prime factorization of 24.
Solve It
2. Write the prime factorization of 100.
Solve It
3. Write the prime factorization of 256.
Solve It
4. Write the prime factorization of 3,872.
Solve It
Factoring Out the Greatest Common Factor
The first line of attack (and the easiest to perform) when factoring an expression is to look for the greatest common factor (GCF), or the largest factor that will divide all terms in the expression evenly. You want a factor that divides each of the terms and, at the same time, doesn’t leave any common factor in the resulting terms. The goal is to take out as many factors as possible. More than one step may be required to completely accomplish this feat, but the end result is a GCF times terms that are relatively prime (have no factor in common).
Q. Find the GCF and factor the expression. 30x4y2 – 20x5y3 + 50x6y.
A. 10x4y (3y – 2xy2 + 5x2). If you divide each term by the greatest common factor, which is 10x4y, and put the results of the divisions in parentheses, the factored form is 30x4y2 – 20x5y3 + 50x6y = 10x4y (3y – 2xy2 + 5x2). It’s like doing this division, with each fraction reducing to become a term in the parentheses:
Q. Factor out the GCF: .
A. . Dealing with fractional exponents can be tricky. Just remember that the same rules apply to fractional exponents as with whole numbers. You subtract the exponents.
Algebra I Workbook For Dummies Page 8