15
COMMON FACTOR
TO FACTOR A NUMBER or an expression, means to write it as a product of factors.
Example 1. Factor 30.
Solution. 30 = 2· 15 = 2· 3· 5
If we begin 30 = 5· 6, we still obtain -- apart from the order -- 30 = 5· 2· 3.
Problem 1. Factor 50.
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Do the problem yourself first!
50 = 2· 25 = 2· 5· 5
Factoring, then, is the reverse of multiplying. When we multiply out, we write
2(a + b) = 2a + 2b.
But if we switch sides and write
2a + 2b = 2(a + b),
then we have factored 2a + 2b as the product 2(a + b).
In the sum 2a + 2b, 2 is a common factor of each term. It is a factor of 2a, and it is a factor of 2b. This Lesson is concerned exclusively with recognizing common factors.
Problem 2. Factor 3x − 3y.
3x − 3y = 3(x − y)
Problem 3. Rewrite each of the following as the product of 2x and another factor.
For example, 10x3 = 2x· 5x².
| a) | 8x = 2x· 4 | b) | 6ax = 2x· 3a | c) | 2x² = 2x· x | ||
| d) | 2x3 = 2x· x² | e) | 4x10 = 2x· 2x9 | f) | 6x5 = 2x· 3x4 | ||
| g) | 2ax6 = 2x· ax5 | h) | 2x = 2x· 1 | ||||
Example 2. Factor 10a − 15b + 5.
Solution. 5 is a common factor of each term. Display it on the left of the parentheses:
10a − 15b + 5 = 5(2a − 3b + 1)
If we multiply the right-hand side, we will get the left-hand side. In that way, the student can always check factoring.
Also, the sum on the left has three terms. Therefore, the sum in parentheses must also have three terms -- and it should have no common factors.
Problem 4. Factor each sum. Pick out the common factor. Check your answer.
| a) | 4x + 6y = 2(2x + 3y) | b) | 6x − 6 = 6(x − 1) | |
| c) | 8x + 12y − 16z = 4(2x + 3y − 4z) | d) | 12x + 3 = 3(4x + 1) | |
| e) | 18x − 30 = 6(3x − 5) | f) | 2x + ax = x(2 + a) | |
| g) | x² + 4x = x(x + 4) | h) | 8x² − 4x = 4x(2x − 1) | |
Problem 5. Factor each sum.
a) 2 + 6 + 10 + 14 + 18 = 2(1 + 3 + 5 + 7 + 9)
b) 30 + 45 + 60 + 75 = 15(2 + 3 + 4 + 5)
Again, the number of terms in parentheses must equal the number of terms on the left . And the terms in parentheses should have no common factors.
Polynomials
A monomial in x is a single term that looks like this: axn, where n is a whole number. The following are monomials in x:
5x8, −3x², 6.
(We say that the number 6 is a monomial in x, because as we will see in Lesson 21, 6 = 6x0 = 6· 1.)
The whole numbers are the non-negative integers: 0, 1, 2, 3, 4, etc.
A polynomial in x is a sum of monomials in x.
5x4 − 7x3 + 4x² + 3x − 2
(For a more complete definition of a polynomial, see Topic 6 of Precalculus.)
The degree of a polynomial is the highest exponent. This is a polynomial of the 4th degree.
The constant term is the term in which the variable does not appear. In other words, it is the number at the end. In this example, the constant term is −2.
(It is called the constant term, because since it does not depend on the variable, the value of that term never changes.)
Problem 6. Describe each polynomial in terms of the variable it is "in," and say its degree.
a) x3 − 2x² − 3x − 4 A polynomial in x of the 3rd degree.
b) 3y² + 2y + 1 A polynomial in y of the 2nd degree.
c) x + 2 A polynomial in x of the 1st degree.
d) z5 A polynomial in z of the 5th degree.
e) 4w − 8 A polynomial in w of the 1st degree.
Factoring polynomials
If every term is a power of x, as in this example,
x7 + 3x6 + 2x5 + x4
then the lowest power is the highest common factor.
x7 +3x6 + 2x5 + x4 = x4(x3 + 3x² + 2x + 1)
For, lower powers are factors of higher powers .
| x7 | = | x4· x3 |
| x6 | = | x4· x2 |
| x5 | = | x4· x |
The lowest power, x4 in this example, typically appears on the right. For when we write a polynomial, we begin with the highest exponent and then descend to the lowest. 7, 6, 5, 4.
Problem 7. Factor these polynomials. Pick out the highest common factor.
a) x8 + x7 + x6 + x5 = x5(x3 + x² + x + 1)
b) 5x5 − 4x4 + 3x3 = x3(5x² − 4x + 3)
c) x3 + x² = x²(x + 1)
d) 6x5 + 2x3 = 2x3(3x² + 1)
e) 2x3 − 4x² + x = x(2x² − 4x + 1)
f) 3x6 − 2x5 + 4x4 − 6x² = x²(3x4 − 2x3 + 4x² − 6)
Problem 8. Factor each polynomial. Pick out the highest common numerical factor and the highest common literal factor.
a) 12x² + 24x − 30 = 6(2x² + 4x − 5).
There is no common literal factor. The sum in parentheses has no common factors.
b) 16x5 − 32x4 + 24x3 = 8x3(2x² − 4x + 3)
c) 36y15 − 27y10 − 18y5 = 9y5(4y10 − 3y5 − 2)
d) 8z² − 12z + 20 = 4(2z² − 3z + 5)
e) 16x² − 24x + 40 = 8(2x² − 3x + 5)
f) 20x4 − 12x3 + 36x² − 4x = 4x(5x3 − 3x² + 9x − 1)
g) 18x8 − 81x6 + 27x4 − 45x² = 9x²(2x6 − 9x4 + 3x² − 5)
h) 12x10 − 6x3 + 3 = 3(4x10 − 2x3 + 1)
Example 3. Factor x²y3z4 + x4yz3.
Solution. The highest common factor (HCF) will contain the lowest power of each letter. The HCF is x²yz3. With that as the common factor, reconstruct each term:
x²y3z4 + x4yz3 = x²yz3(y²z + x²)
If we multiply the right-hand side, we will obtain the left-hand side.
Problem 9. Factor.
a) 3abc− 4ab = ab(3c − 4)
b) 2xy − 8xyz = 2xy(1 − 4z)
c) x²y3 − x3y² = x²y²(y − x)
d) 8ab3 + 12a²b² = 4ab²(2b + 3a)
e) a5b5 − a8b² = a5b²(b3 − a3)
f) x6yz² + x²y4z3 − x3y3z4 = x²yz²(x4 + y3z − xy²z²)
Next Lesson: Quadratic Trinomials
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