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 Dependent Variable

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# Factoring Polynomials and Solving Higher Degree Equations

Example 5. The polynomial p(x) = x2 + 1 is irreducible.

Justification. p(x) is quadratic. Therefore if it has a non constant factor, this factor has to be
linear. Now every linear polynomial has a root. In sum, if p(x) was reducible then it would have
a root. But, p(x) does not have roots, because no matter what real number we substitute for x,
x2 will be greater or equal than 0, and adding one will make x2 + 1 greater than 0. So, p(x) will
always evaluate to a positive number.

In sum, p(x) has no roots and therefore it is irreducible.

The previous example illustrates a very important test for deciding whether a quadratic polynomial
is irreducible.

 A quadratic polynomial is irreducible exactly when it has no roots .

Use this test to justify each of the following true statements:

1. The polynomial x2 + 7 is irreducible.

2. The polynomial −2x2 − 5 is irreducible.

3. The polynomial (x − 2)2 + 21 is irreducible.

4. The polynomial x2 − 9 is reducible.

5. The polynomial (x − 2)2 + (x + 3)2 is irreducible.

Identifying common factors

The most straightforward method for factoring is identifying common factors among the terms of
a polynomial. Then we can use the distributive property, in the contracting4 direction. Let’s start
with some examples:

Example 6. Factor the polynomial: ax + bx.

Answer. Each of the terms of the polynomial that we have to factor has x as factor. So we can use
the distributive property to get:

ax + bx = (a + b)x

Example 7. Factor: 2x2 − 4x3.

Answer. In this case the coefficients have 2 as a common factor and the variable parts of the two
terms have x2 as a common factor. So 2x2 is a common factor of the two terms, and we have:

2x2 − 4x3 = 2x2(1 − 2x)

In general to identify a possible common factor among the terms of a polynomial we can follow
the following procedure:

1. Check if there is a common factor of all the coefficients. If there is such a common factor, it
will be the coefficient of the common factor.

2. For each variable of the polynomial, check whether it occurs in all the terms. Each variable
that occurs in all the terms will occur in the common factor, and its exponent in the common
factor will be the smallest of the exponents in all occurrences.

Once the common factor has been identified we can find the other factor in the factorization of the
polynomial we proceed as follows.

3. We divide every term of the original polynomial by the common factor. The quotient will be
a term of the other factor.

Example 8. Factor the polynomial p(x, y, z,w) = 3x2y3z − 6xy2z2 + 9x3y2w.

Answer. The coefficient of the common factor will be 3.

x occurs in the first term with exponent 2, in the second term with exponent 1 and in the third
term with exponent 3. So the common factor will have an x.

y occurs with exponent 3 in the first term, 2 in the second and 2 in the third. So the common
factor will have a y2.

z does not occur in the third term. So z doesn’t occur in the common factor.

w does not occur in the first (or the second) term. So w doesn’t occur in the common factor.
In sum the common factor is 3xy2.

Now we divide each term of p(x, y, z,w) with the common factor to get the terms of the other
factor:

3xy2(xyz − 2z2 + 3x2w)

Let’s practice:

1. Factor 6x4y3z − 12x2yz3 + 21x3yz2.

2. Factor 4xy2 − x2y3 + 8x3y4.

3. Factor 2x2y3 − 3x3z4 + 5yz2.

4. Factor 6yw3x3 − 24w3x3y3 − 12w4x2.

5. Factor −7x4 − 14x3 + 21x2 + 7x.

The technique of identifying common factors can be used any time that we have a polynomial
written as a sum of products, even if the polynomial is not in simplified expanded form.

Example 9. Factor 2x(x + 3) − 7x5(x + 3).

Answer. Both summands have x + 3 as a factor. So we can write:

2(x + 3) − 7x5(x + 3) = (x + 3)(2 − 7x5)

Example 10. Factor (2x − 3)xy2 − 7(2x − 3)x2.

Answer. Now we have x(2x − 3) as a common factor. So:

(2x − 3)xy2 − 7(2x − 3)x2 = x(2x − 3)(y2 − 7x2)

Often the common factor is “in disguise”, and we need to look carefully to be able to identify
it. For example, the common factor may appear with opposite signs in different summands:

Example 11. Factor x2(3x + 1) + 4y(−3x − 1).

Answer. In this case (3x + 1) is a common factor, since −3x − 1 = −(3x + 1). So:

x2(3x + 1) + 4y(−3x − 1) = (3x + 1)(x2 − 4y)

Example 12. Factor 3x(x − 2) + 5y(2 − x).

Answer. x − 2 is the common factor:

3x(x − 2) − 5y(2 − x) = (x − 2)(3x + 5y)

Another “disguise” to watch for, is the lack of parenthesis, sometimes the common factor may
appear by itself: