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.
To see the answer, pass your mouse over the colored area.
50 = 2· 25 = 2· 5· 5
Factoring, then, is the reverse of multiplying. When we multiply, 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 this 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. Rule 1 of exponents.
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)
You can always check factoring by multiplying the right-hand side. It should produce the left-hand side.
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.
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.
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.)
A polynomial in x is a sum of monomials in x.
5x4 − 7x3 + 4x2 + 3x − 2
When we write a polynomial, the style is to begin with the highest exponent and go to the lowest. 4, 3, 2, 1.
(For a more complete definition of a polynomial, see Topic 6 of Precalculus.)
The degree of a polynomial is the highest exponent. The polynomial above is of the 4th degree.
(We call it the constant term because even when the value of the variable changes, the value of the constant term does not change. It is constant.)
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.
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 .
The lowest power, x4 in this example, typically appears on the right. Again, when we write a polynomial, we begin with the highest exponent and go to the lowest. 7, 6, 5, 4.
Once more, to say that we have factored the polynomial on the left --
x7 +3x6 + 2x5 + x4 = x4(x3 + 3x2 + 2x + 1)
-- means that we will obtain that polynomial if we multiply the factors on the right.
The student should confirm that.
Problem 7. Factor these polynomials. Pick out the highest common factor.
(How can you check your factoring? By multiplying!)
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 you multiply the right-hand side, you 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)
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