FACTORING POLYNOMIALS

Objective:   To recall how polynomials factor.

Recall 1: We recall that

x^2 + 5*x = x*(x + 5),

that

x^3 + 5x^2 + 4x = x(x^2 + 5x + 4),
and that

3x^3 + 6x^2 + 9x = 3x(x^2 + 2x + 3).

Recall 2: We recall that

(x + 5)  (x - 5) = x^2 - 25,

and that, more generally,

(x + b)(x - b) = x^2 - b^2.

Recall 3: A slightly different form of this last recall is

(3x + 5)(3x - 5) = 9x^2 - 25,

or, more generally,

(ax + by) (ax - by) = (ax)^2 - (by)^2.

Notice that if a = 1,   and y = 1, this reduces to the form in Recall 2.

Note

Very often, a difference of two squares is not recognized, simply because it is not written in a recognizable form.
For example, 6x^3 - 150x^2 does not look like the difference of two squares.  Once you factor out a copy of 6x from both terms, you have the expression 6x(x^2 - 25).  The term in parentheses is the difference of two squares.

Recall 4: Look for all the products that repeatedly appear in each  term of a polynomial expression.  For example, the expression

5x^2*y^2 + 3xy^3 +xy^2

is the same as

5x(xy^2) + 3(xy^2)y + (xy^2).

Since an (xy^2) appears in each term, we may factor it out and rewrite it  as

(xy^2)(5x + 3y +1).

Recall 5: Another product to recall is

(x - 1)  (x^2 + x + 1) = x^3 - 1,

and that, more generally,

(x - b) (x^2 + bx + b^2) = x^3 - b^3.

The expression on the right is occasionally referred to as the difference between two cubes.

Note
Very often, a difference of two cubes is not recognized, simply because it is not written in a recognizable form.
For example, 4x^4 - 108x does not look like the difference of two squares.  Once you factor out a copy of 4x from both terms, you have the expression 4x(x^3 - 27).  The term in parentheses is the difference of two cubes.

Examples

Example 1. Factor  55x^22 + 11x^11.

Solution: Notice that there is a copy of 11x^11 in 55x^22, since

55x^22 = (11x^11)(5x^11).

So the factored form of the given polynomial is 11x^11(5x^11 + 1).

Note
Always look for the lowest common power of the variable that appears in all the terms of the polynomial.

Example 2. Factor 8x^3 - 200x.

Solution: There is a copy of 8x in each term.  Notice that

8x^3 = (8x)x^2, and that

-200x = (8x)(-25).

Therefore, we may pull out a copy of 8x from each term to rewrite  the original expression as

8x(x^2 - 25).

Now, notice that x^2 - 25 is the difference between two squares and may  be factored as
(x + 5)(x - 5);

so the original expression may be completely factored as

8x(x + 5)(x - 5).

Example 3. Factor x^2 + 24x + 143.

Solution: Look for the factors of 143.

143 = 1 *143,
= (-1) * (-143),
= 11 * 13,
= (-11) * (-13).

The only factors whose sum is +24 are 11 and 13; and so we have:

x^2 + 24x + 143 = (x + 11)(x + 24).

Note
In general, to factor a polynomial such as x^2 + bx + c, look for products of the form

(x ± some number)*(x ± some number).

Sometines you may not be able to factor the polynomial in this way.  We shall consider this situation in Workout 13.

Example 4 Factor  55x^100*y^50*z^25 + 110x^25*y^50*z^100.

Solution: Write 110 as 55 * 2, so the expression becomes

55(x^100*y^50*z^25 + 2*x^25*y^50*z^100).

Next, notice that the lowest common power of x in both terms is 25,  the lowest common power of y in both terms is 50 and that the lowest  common power of z in both terms is 25.  So, we may factor out the  product of all these lowest common powers, namely x^25*y^50*z^25.  We  may, therefore, rewrite the expression as

55x^25*y^50*z^25.(x^75 + 2z^75).

Example 5.  Factor x^3 - 1331.

Solution: This is the difference of two cubes, since 11^3 = 1331.  We may use the formula of the difference between two cubes:

x^3 - a^3 = (x - a)(x^2 + ax + a^2)

In the case of this example, a^3 = 1331, so a = 11.  Therefore,

x^3 - 1331= (x - 11)(x^2 + 11x + 11^2)
= (x - 11)(x^2 + 11x + 121).

Note

If the example above asked you to factor x^3 + 1331, then you could write it as x^3 - (-1331), which is the same as x^3 - (-11)^3.  You would then have a factoring like
x^3 + 1331= (x - (-11))(x^2 + (-11)x + (-11)^2)
= (x + 11)(x^2 - 11x + 121).

Example 6 Show that 6x^4 - 1296x = 6x(x - 6)(x^2 + 6x + 36).

Solution:  Occasionally, differences of two cubes disguise themselves.  In this case, we may remove the disguise by factoring out a copy of 6x from both terms to get

6x(x^3 - 216).

The remaining part of the demonstration follows from an argument similar to the previous example; i.e. that

(x^3 - 216) = (x - 6)(x^2 + 6x + 36), since 216 = 6^3.

EXERCISES

Factor the following polynomials:

1. x^2 + x

2. x^3 + x^2 + x

3.  4x^3 - 3x^2 + 5x

4. x^2 - 36

5. 4x^2 - 100

6. 3x^3 - 75x

7. x^2 + 6x + 9

8. x^2 - 8x + 16

9. x^2 + 3x - 10

10. x^3 + 7x^2 + 12x

11. x^3*y - y^3*x

12. 5x^2y^2 + 3xy^3 +xy^2

13. sqrt(5)*x^3y + pi*y*x^2

14. 36 + 12*x + x^2

15. x^3 - 27

16. x^4 - 27*x

1. x*(x + 1)

2. x*(x^2 + x + 1)

3  x*(4*x^2 - 3*x + 5)

4 (x - 6)*(x + 6)

5 4*(x - 5)*(x + 5)

6 3*x*(x - 5)*(x + 5)

7.  (x + 3)*(x + 3)

8 (x - 4)*(x - 4)

9 (x + 5)*(x - 2)

10. x*(x + 3)*(x + 4)

11 x*y*(x - y)*(x + y)

12 x*y^2*(5*x + 3*y + 1)

13 x^2*y*(*x + )

14 (x + 6)*(x + 6)

15 (x - 3)*(x^2 + 3*x + 9)

16 x*(x - 3)*(x^2 + 3*x + 9)