Factoring Trinomials

We use three different techniques to factor trinomials depending on preference and the numbers we have to factor.

For instance, if we want to factor an expression such as

x² – 4x – 5

we would use the trial and error or inspection method since
there's very little chance of going wrong because
both 1, (the coefficient of x²) and –5, (the last or constant term) are prime numbers.

So x² can only be x times x, and –5 can only be –5 times 1 or –1 times 5.

If however, we have to factor an expression such as 15a ² – 29ab – 14b ², it would be inefficient to use trial and error since 15, (the coefficient of a²) and 14, (the coefficient of b²) can both be factored in 2 ways. 15 can be 15 × 1 or 5 × 3, and 14 can be 14 × 1 or 2 × 7, so it could take time to find the correct combination of factors that will produce the given trinomial with the correct middle term.

In cases such as this, we use either the magic number technique or the factor theorem to factor the given trinomial.

Perfect Square Trinomial

A perfect square trinomial is the result of squaring a binomial.

Perfect square trinomials are easily recognized because their first and last terms are perfect squares and the middle term is twice the product of their square roots.

For example, x² – 6x + 9 is the perfect square of (x – 3) because the first and last terms, x² and 9 are perfect squares, and – 6x is 2 times x (the root of x²)
times –3 (a square root of 9).

If we'd had x² – 6x – 9, it would not be a perfect square
since –9 is not a perfect square (in the Real numbers).

hint: for a trinomial to be the perfect square of a binomial, the sign in front of the first and last terms must be + (positive) since the square of any term (positive or negative) is positive. (in the Real numbers)

Some examples of perfect square trinomials

examples

Factor completely

a) x² – 6x + 9 – y² =	a) 1st 3 terms are a perfect square trinomial
(x – 3)² – y² =	difference of squares
[(x – 3) + y] [(x – 3) – y] =	(sum of roots) (difference of roots)
(x – 3 + y) (x – 3 – y)	remove brackets
b) 4a² + 8a + 4 – 9b² =	b) common factor of 4 in 1st 3 terms
4(a² + 2a + 1) – 9b² =	4(perfect square trinomial) – square of 3b
4(a² + 2a + 1) – 9b² =	difference of squares
[2(a + 1) + 3b][2(a + 1) – 3b] =	(sum of roots) (difference of roots)
(2a + 2 + 3b) (2a + 2 –3b)	remove brackets
c) 25m² – p² + 14p – 49 =	c) common factor of –1 in last 3 terms
25m² – (p² – 14p + 49) =	last 3 terms are a perfect square trinomial
25m² – (p – 7)² =	difference of squares
[5m + (p – 7)][5m – (p – 7)] =	(sum of roots) (difference of roots)
(5m + p – 7) (5m – p + 7)	remove brackets

trial and error or inspection method

If our trinomial has small or prime number coefficients like 2, or 3, we can usually find the right combination of factors by inspection if we follow a few simple rules.

1. Remember we cannot change the value of the terms, we change only the form.
   The product of our factors must be equal to the original trinomial. When we find the product of our factors, we get exactly what we started with.
2. The middle term must be the sum of (Outer × Outer) + (Inner × Inner).
3. If the 3rd term is positive, both brackets have the same sign as that of the middle term, since a positive product is the result of multiplying 2 negatives or 2 positives.
4. If the 3rd term is negative, the brackets have opposite signs,
   one positive, one negative.
   The larger of the (Outer × Outer) and (Inner × Inner) products will have the same sign as the middle term.

   Sounds complicated -- but it'll be clear when we look at some examples.

Examples

Factor completely

a) x² + 5x + 4 =
( x + ) ( x + ) =	both middle and last terms are +, both brackets must have +
(x + 4) ( x + 1)	factors of 4 which add to 5 are 4 and 1.
b) x² – 5x + 4 =
( x – ) ( x – ) =	+ with the last term, – with the middle term , both brackets –
( x – 4 ) ( x – 1)	The factors of 4 which add to –5 are –4 and –1.
c) x² – 3x – 4 =
( x – ) ( x + ) =	– with the last term and middle, so the brackets signs opposite
( x – 4 ) ( x + 1)	The factors of –4 which add to –3 are –4 and +1.
d) 2x² + 7x + 6 =	Both middle & last terms +, so brackets have +
(2x + __ )(x + __ ) =	We need factors of 6 that give a middle term of 7
(2x + 3 )(x + 2 )	O × O = 4x & I × I = 3x so the factors are as shown

Magic Number Method

This method involves breaking up the trinomial's middle term into 2 terms such that there is a common factor in the 1st and 2nd terms as well as in the 3rd and 4th terms. We then use a common factor grouping approach to factor the trinomial which we've rewritten as 4 terms.
To illustrate, let's do the same examples again using this method.

Factor completely

a) x² + 5x + 4 = x² + x + 4x + 4 = x( x + 1) + 4( x + 1) = ( x + 1 ) ( x + 4 ) b) x² – 5x + 4 = x² – x – 4x + 4 = x( x – 1 ) – 4 ( x – 1) = ( x – 1 ) ( x – 4 ) c) x² – 3x – 4 = x² + x – 4x – 4 = x( x + 1) – 4( x + 1) = ( x + 1 ) ( x – 4 ) d) 2x² + 7x + 6 = 2x² + 4x + 3x + 6 = 2x(x + 2) + 3(x + 2) = ( x + 2 )( 2x + 3 )

a) We want to write the middle term, 5x, as two terms which have common factors with x² and + 4 as shown. Now we group factor the (x + 1) to get exactly what we had before. b) We want to write the middle term, –5x, as two terms which have common factors with x² and +4 as shown. Now take out the common factor of (x – 1) and get exactly what we had before. c) We want to write the middle term, –3x, as two terms which have common factors with x² and –4 as shown. Now take out the common factor of (x + 1) and get exactly what we had before. d) We want to write the middle term, +7x, as two terms which have common factors with 2x² and +6 as shown. Now take out the common factor of (x + 2) and get exactly what we had before.

Factor Theorem Method

The theorem states: if substituting x = a makes an expression = 0,
then (x – a) is a factor of that expression.

To illustrate, pick a = 1 and set x = a. Then x² – 2x + 1 = 0, [(1)² – 2(1) + 1 = 0] .

So, ( x – 1 ) which is x – a is a factor of x² – 2x + 1.

Dividing x² – 2x + 1 by ( x – 1 ), its known factor, we find the other factors.

The problem with this method is that dividing a trinomial by a binomial can be difficult at times. Also, it is often hard to find a number to make the expression = 0.

When using the factor theorem, we try factors of the third term "a", however, some numbers
such as 64, have a boatload of factors, so it could take a long time to find the right ones.

To divide a trinomial by a binomial: (trinomial = dividend = dv, binomial = divisor = ds )

1. Find the quotient (qt) of the first term of the trinomial & the first term of the binomial.
2. Multiply the binomial by this quotient, subtract this product from the trinomial.
3. Now find the quotient of the first term of the remainder divided by the first term of the binomial.
4. Multiply the binomial by this quotient and subtract the product from the remainder.

To illustrate, let's divide x² – x – 2 by (x + 1) which is (x – a) if a = –1.

1. Divide x² (dv) by x (1st term in ds) to get x. Write this x up top in the quotient (qt).
2. Multiply (x + 1), the divisor, by the quotient, (the x up top), to get x² + x (line 3)
3. Subtract this x² + x (ln 3) from x² – x (dv) to get – 2x. Bring down – 2 from (dv) (line 4)
4. Divide – 2x – 2 (line 4) by the divisor (x + 1) to get – 2. Write this – 2 up top in quotient.
5. Multiply (x + 1), the divisor, by the – 2 in quotient to get – 2x – 2 (line 5)
6. Subtract to get 0 -- no remainder.

   

   So the factored form of x² – x – 2 is (x + 1)(x – 2).

   examples:

Factor completely:

a) x² – 5x + 6 = (2)² – 5(2) + 6 = 0 (x² – 5x + 6) ÷ (x – 2) = x – 3 x² – 5x + 6 = (x – 2)(x – 3) b) 3x² + 2x – 1= 3(–1)² + 2(–1) –1 = 0 (3x² + 2x – 1) ÷ (x + 1) = 3x – 1 3x² + 2x – 1 = (x + 1) ( 3x – 1)

a)Set x = factors of 6 which make x² – 5x + 6 = 0. Try x = 2. It works, so (x – 2) is a factor. Divide x² – 5x + 6 by (x – 2) to get (x – 3). The factors of x² – 5x + 6 are (x – 2) and (x–3). b) The factors of –1 are 1 and –1. Set x = –1. It works, so [x – (–1)] is a factor. Divide 3x² + 2x – 1 by (x + 1) to get (3x – 1). The factors of 3x² + 2x – 1 are (x + 1) and (3x – 1).

.

Note: of the three methods for factoring trinomials, the inspection method is the most commonly used and with some practice, you can get good at it. Just remember to check the value of the middle term to verify that you have chosen the right combination of factors

practice:

factor completely

a) tx² – 4txy + 4txy²	(b) 2m² – 6m + mn – 3n	(c) 9x² + 6xy + y² – 16
d) 162 – 2m4 .	e) a²x² – x² – a² + 1	f) 24c4 – 81cd3
g) 16r² – 4s² + 28st – 49t²	h) 128x6 – 2y6	i) 9a ² + 4bc – 4c ² – b ²
j) 6y² + 11y + 3	k) 6 – 19bc + 3b²c²	l) 3t ² – 22t + 7

.

solutions:

a) tx (x – 4y + 4y ² )	b)2m ( m – 3 ) + n ( m – 3 ) = (m – 3 ) ( 2m + n )
c) ( 3x + y ) ² – 4 ² = ( 3x + y + 4 ) ( 3x + y – 4 )	d) 2 ( 81 – m4 ) = 2 ( 9 + m² ) ( 9 – m² ) 2 (9 + m² ) (3 – m) (3 + m)
e) x² ( a² – 1 ) – ( a² – 1 ) = ( x² – 1 ) ( a² – 1 ) = ( x + 1 ) ( x – 1 ) ( a + 1 ) ( a – 1 )	f) 3c (8c3 – 27d3 ) = 3c(2c – 3d )(4c² + 6cd + 9d²)
g)16r² – (4s² – 28st + 49t²) = (4r)² – (2s – 7t )² = [4r + (2s – 7t ) ] [4r – (2s – 7t ) ] = ( 4r + 2s – 7t ) (4r – 2s + 7t )	h) 2( 64x6 – y6 ) = 2( 8x3 – y3 ) (8x3 + y 3) = 2(2x–y)(4x² + 2xy+ y²)(2x + y)(4x² – 2xy+ y²)
i) 9a² – ( b² – 4bc + 4c² ) = (3a)² – ( b – 2c )² = [ 3a + ( b – 2c )] [ 3a – ( b – 2c ) ] = ( 3a + b – 2c ) (3a – b + 2c )	j) ( 3y + 1 ) ( 2y + 3 ) k) ( 6 – bc ) ( 1 – 3bc ) l) ( 3t – 1 ) ( t – 7 )

.

Factoring Summary

1. Common Factor: Take out the common factor x ² y³ – 3xy ² = xy ² (xy – 3)
2. Difference of Squares: x² – y² = (x + y)(x – y)
3. Sum or Difference of Cubes: 2 brackets, 1st cube roots & same sign.
   2nd bracket from terms in first bracket.
   Square first term, multiply terms & change sign, square last term.
   27x ³ – y ³ = (3x – y)(9x ² + 3xy + y ²), x ³ + 64y ³ = (x + 4y)(x² – 4xy + 16y ²)
4. Perfect Square Trinomials: perfect squares for 1st and last terms, middle term is double the product of the roots of the first and last terms.
5. Trinomial Factoring: Use either inspection, magic number, or factor theorem method. Check that the middle term is correct!!

   intro	perfect squares	inspection method	magic number
   factor thm.	practice	solutions	factoring summary

Algebra MathRoom Index

.

(all content © MathRoom Learning Service; 2004 - ).