trinomial factoring |
We use three different techniques to factor trinomials depending on preference and the numbers we have to factor.
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.
intro | perfect squares | inspection method | magic number |
factor thm. | practice | solutions | factoring summary |
A trinomial is an algebraic expression with 3 terms. ax² + bx + c is a 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).
Some examples of perfect square trinomials
4x² + 20xy + 25y² = (2x + 5y)² | 9a² – 24a + 16 = (3a – 4)² |
m² – 6mn + 9n² = (m – 3n)² | y² + 18y + 81 = (y + 9)² |
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 |
intro | perfect squares | inspection method | magic number |
factor thm. | practice | solutions | factoring summary |
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.
Sounds complicated -- but it'll be clear when we look at some 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 |
intro | perfect squares | inspection method | magic number |
factor thm. | practice | solutions | factoring summary |
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. |
intro | perfect squares | inspection method | magic number |
factor thm. | practice | solutions | factoring summary |
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 )
To illustrate, let's divide x² – x – 2 by (x + 1) which is (x – a) if a = –1.
So the factored form of x² – x – 2 is (x + 1)(x – 2).
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). |
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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
intro | perfect squares | inspection method | magic number |
factor thm. | practice | solutions | factoring summary |
factor completely
a) tx² – 4txy + 4txy² | (b) 2m² – 6m + mn – 3n | (c) 9x² + 6xy + y² – 16 |
d) 162 – 2m^{4} . | e) a²x² – x² – a² + 1 | f) 24c^{4} – 81cd^{3} |
g) 16r² – 4s² + 28st – 49t² | h) 128x^{6} – 2y^{6} | i) 9a^{ }² + 4bc – 4c^{ }² – b^{ }² |
j) 6y² + 11y + 3 | k) 6 – 19bc + 3b²c² | l) 3t^{ }² – 22t + 7 |
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intro | perfect squares | inspection method | magic number |
factor thm. | practice | solutions | factoring summary |
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 – m^{4} ) = 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 (8c^{3} – 27d^{3} ) = 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( 64x^{6} – y^{6} ) = 2( 8x^{3} – y^{3} ) (8x^{3} + 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 ) |
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intro | perfect squares | inspection method | magic number |
factor thm. | practice | solutions | factoring summary |
intro | perfect squares | inspection method | magic number |
factor thm. | practice | solutions | factoring summary |
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