A/ FACTORING

Factoring is a basic skill and must be mastered with efficiency. The easiest way to do that is to remember and use the words that describe the different factoring methods we use to turn messy polynomials into pretty products. Once we know that P(x) = (x – a)(x – b)(x – c), we know the zeros of the polynomial and can graph it quite easily. We can also easily locate the maximums and minimums of the function. This lessons covers 5 different factoring methods.

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1) Common Factor:

ex 1a: ax – ay + az = a(x – y + z)

ex 1b: 2bx + 4by – 18bz = 2b(x + 2y – 9z)

Group Common Factor

ex 2a: ax + ay – bx – by = a(x + y) – b(x + y) = (a – b)(x + y)

ex 2b: 8x² + 16y² – x³ – 2xy² = 8(x² + 2y²) – x(x² + 2y²) = (8 – x)(x² + 2y²)

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2) Difference of Squares:

ex 3a: x² – y² = (x + y)(x – y)

difference of squares = (sum of the roots)(difference of the roots)

ex 3b: 4a² – 16b² = 4(a² – 4b²) = 4(a – 2b)(a + 2b) -- (a combo of common and diff of sqs.)

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3) Sum or Difference of Cubes:

ex 4: x³ + y³ = (x + y)(x² – xy + y²)

ex 5: x³ – y³ = (x – y)(x² + xy + y²)

n.b.: to factor a sum or difference of cubes we make 2 brackets.
in the first: put the cube roots and the same sign
in the 2nd: square the 1st term, multiply the terms and change the sign,
square the last term in the first bracket.

Examples:

a) sum of cubes 27c³ + 8d³ = (3c + 2d)(9c² – 6cd + 4d²)	b) difference of cubes 64 – 125x³ (4 – 5x)(16 + 20x + 25x²)
c) difference of squares, then sum and difference of cubes. x 6 – 64y 6 (x3 – 8y3) (x3 + 8y3) = (x – 2y)(x² + 2xy + 4y²)(x + 2y)(x² – 2xy + 4y²)

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4) Trinomials:

There are 3 types of trinomials: perfect square, simple and complex trinomials.

A perfect square trinomial is the square of a binomial.

x² – 8x + 16 = (x – 4)² and 4x² + 12x + 9 = (2x + 3)²

In a simple trinomial the coefficient of x² is always equal to 1.

In a complex trinomial the coefficient of x² is not equal to 1.

Perfect squares can be either simple or complex trinomials.

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4a) Perfect Square Trinomials:

The 1st and last terms are perfect squares and the middle term is double the product of the roots of those perfect squares. Since all perfect squares are positive in the set of Real numbers, the sign before the 3rd term will always be +. The sign of the middle term tells us if the binomial is a sum or difference.

ex 6: x² + 6xy + 9y ² = (x + 3y)(x + 3y) = (x + 3y)² : the binomial is a sum.

but: 4x ² – 20xy + 25y ² = (2x – 5y)² : the binomial is a difference.

ex 7a: Here, the first 3 terms are a perfect square binomial.

x ² + 6xy + 9y ² – 16 = (x + 3y)² – 4² = (x + 3y + 4)(x + 3y – 4)

Then it's a difference of squares.

ex 7b: Here, the last 3 terms are a perfect square binomial when we factor out – 1.

4a ² – 9b ² – 24b – 16 = 4a ² – (9b ² + 24b + 16) = 4a ² – (3b + 4)²

we get [2a + (3b + 4)] [2a – (3b + 4)]

n.b. when the perfect square trinomial is after the minus sign, use inner brackets in your expression of the factors.

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4b/ Simple Trinomials

The coefficient of the squared term is always = 1

example 8: x² – 5x + 6 = (x – 3)(x – 2) we need factors of 6 that add to 5, both are negative.

example 9: x² + 5x + 6 = (x + 3)(x + 2) we need factors of 6 that add to 5, both are positive.

example 10: x ² – x – 6 = (x – 3)(x + 2) we need factors of 6 that subtract to 1, the bigger one is negative.

Notes:

if the 3rd term's sign is positive, both brackets have the same sign as the middle term. The inners and outers ADD to give the middle term.

If the 3rd term's sign is negative, the brackets have opposite signs. The inner and outer products SUBTRACT to give the middle term.

4c/ Complex Trinomials

The coefficient of the squared term is not 1.
Watch for prime numbers, they're easy to factor.

example 11: 2x² – 5x + 2 = (2x – 1)(x – 2) outers + inners = – 5x we get – 4x from outers, – x from inners, both are negative.

example 12: 3x² + 11x + 6 = (3x + 2)(x + 3) outers + inners = 11x we get 9x from outers, 2x from inners, both are positive.

example 13: 5x ² – 2x – 3 = (5x + 3)(x – 1) difference of outers & inners subtract to 2, the bigger product (outers) is negative.

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5/ The Factor Theorem

When we need to factor a polynomial of degree 3 or higher, we use the factor theorem. It says if we can find a value for x that makes the polynomial equal 0, then x minus that value is a factor of the polynomial. So, if x = a makes P(x) = 0, then (x – a) is a factor of P(x). The other factors can then be found by dividing P(x) by (x – a). It's pretty obvious that a must be a factor of the constant term in the polynomial.

The Factor Theorem If P(x) = a0 + a1 x + a2 x ² + ..... + an x n and P(a) = 0, then (x – a) is a factor of P(x) and P(x) = (x – a) [g(x)]

example 14: factor 3x ³ + 7x² – 2x – 8

If we set a = – 2 the expression = 0. This means that it's divisible by (x + 2).

Now the quotient, 3x² + x – 4 = (3x + 4)(x – 1)

so, 3x³ + 7x² – 2x – 8 = (x + 2)(3x + 4)(x – 1)

Note: if P(x) is missing powers, make sure to enter 0x ⁿ where n is a missing power when dividing,. For example, 3 x ³ – 2x – 8 will become 3x³ + 0x² – 2x – 8 so that we can line up the like terms that result from multiplying divisor by quotient.

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EXERCISE 1:

Factor completely:

1) cd + cdx	2) x³ + 3x² + x	3) tx² – 4txy + 4txy²
4) 2m ² – 6m + mn – 3n	5) 9x ² + 6xy + y ² – 16	6) b4 – 16
7) 81 – m4	8) a²x ² – x ² – a ² + 1	9) 9a2 – 12ab + 4b²
10) 6 – 19bc + 3b2c²	11) 6y2 + 11y + 3	12) 3t2 – 22t + 7
13) x3 – 27y³	14) 24c4 – 81cd³	15) 64 – x6
16) 9 – x2 – 2xy – y²	17) 16r2 + 28st – 4s2 – 49t²	18) 9a2 + 4bc – 4c2 – b²
	19) 30xy – 16z2 + 9x2 + 25y²

20) Use the factor theorem to find the zeros of these polynomials.

a) P(x) = x³ + 4x² + x – 6

b) P(x) = x³ + 5x² – 2x – 24

c) P(x) = x³ – 6x² + 3x + 10

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Solutions for Exercise 1

1) cd(1 + x)	2) x(x² + 3x + 1)	3) tx(x – 4y + 4y²)
4) (2m + n) (m – 3)	5) (3x + y + 4)(3x + y – 4)	6) (b – 2)(b + 2)(b² + 4)
7) (3 – m)(3 + m)(9 + m²)	8) (a – 1)(a + 1)(x – 1)(x + 1)	9) (3a – 2b)²
10) (6 – bc)(1 – 3bc)	11) (2y + 3)(3y + 1)	12) (3t – 1)(t – 7)
13) (x – 3y)(x² + 3xy + 9y²)	14) 3c(2c – 3d)(4c² + 6cd + 9d²)
15) (2 – x)(4 + 2x + x²)(2 + x)(4 – 2x + x²)		16) (3 + x + y)(3 – x – y)
17) (4r + 2s – 7t)(4r – 2s + 7t)	18) (3a – b + 2c)(3a + b – 2c)	19) (3x + 5 + 4y)(3x + 5 – 4y)

20) Use the factor theorem to find the zeros of these polynomials.

a) P(1) = 0 so (x – 1) is a factor.

the factors are (x – 1)(x + 3)(x + 2) and the zeros are 1, – 3 and – 2.

b) P(2) = 0 so (x – 2) is a factor.

the factors are (x – 2)(x + 3)(x + 4) and the zeros are 2, – 3 and – 4.

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c) P(2) = 0 so (x – 2) is a factor.

the factors are (x – 2)(x + 1)(x – 5) and the zeros are 2, – 1 and 5.

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