SETASIGNThis e-book
Tutorial 3.5 Factor “special” polynomials through patterning
Mathematicians always look for easier or quicker ways to complete tasks. Thus, some factoring tasks can be completed through the use of previously established patterns. If we know these patterns, we can just present the factors without doing too much back-up work.
When we multiply two identical binomial factors, we get a perfect square trinomial:
(a + b)2 = (a + b)(a + b) = a(a) + a(b) + b(a) + b(b) = a2 + 2ab + b2
(a – b)2 = (a – b)(a – b) = a(a) – a(b) – b(a) – b(–b) = a2 – 2ab + b2
When we multiply two binomials such as (a + b)(a – b), we get a difference of squares binomial:
(a + b)(a – b) = a(a) + a(–b) + b(a) + b(–b) = a2 – b2
NOTE: There is no such thing as the sum of squares binomial.
When we multiply three identical binomial factors, we get a perfect cube trinomial:
(a + b)3 = (a + b)(a + b)(a + b) (a + b) × [(a + b)(a + b)]
(a + b) × [a(a) + a(b) + b(a) + b(b)]
(a + b) × [a2 + 2ab + b2]
a(a2) + a(2ab) + a(b2) + b(a2) + b(2ab) + b(b2) a3 + 2a2b + ab2 + a2b + 2ab2 + b3
a3 + 3a2b + 3ab2 + b3 (a – b)3 = (a – b)(a – b)(a – b) (a – b) × [(a – b)(a – b)]
(a – b) × [a(a) + a(–b) – b(a) – b(–b)]
(a – b) × [a2 – 2ab + b2]
a(a2) – a(2ab) + a(b2) – b(a2) – b(–2ab) – b(b2) a3 – 2a2b + ab2 – a2b + 2ab2 – b3
a3 – 3a2b + 3ab2 – b3
Two other special polynomials are displayed below:
sum of cubes:
a3 + b3 = (a + b)(a2 – ab + b2)
difference of cubes:
a3 – b3 = (a – b)(a2 + ab + b2)
NOTE: If the given binomial is a sum of cubes, then the smaller binomial factor will also be a sum and the middle term of the trinomial factor will be subtracted. However, if the given binomial is a difference of cubes, then the smaller binomial factor will also be a difference and the middle term of the trinomial factor will be added.
When we attempt to factor polynomials through patterning, we should follow these guidelines:
1. Factor out the greatest common monomial factor (if one exists).
2. Determine which of the factoring patterns might apply for the given polynomial.
3. Factor the given polynomial using the proper pattern.
example 3.5a Factor: 9x2; – 12x + 4
Since a = +9 and c = +4 in the given trinomial, we are looking for two numbers that multiply to be (+9)(+4) = +36 and that also add to be –12. Since the 36 is positive, the two desired numbers will have the same sign…and, since the 12 is negative, these signs will be negative. Therefore, the desired numbers are: –6, –6. Since both of these numbers are identical, the given trinomial is a perfect square. Thus, we can use a pattern to complete this task…
given polynomial: 9x2 –12x + 4 = (a – b)2 perfect square pattern: a2 –2ab + b2 = (a – b)2 Using the pattern: a2 = 9x2 = (3x)2
a = 3x
–2ab = –12x –2(3x)b = –12x
(–6x)b = –12x b = 2
b2 = 4 = (2)2 b = 2
Therefore: 9x2 – 12x + 4 = (3x – 2)2 = (3x – 2)(3x – 2)
189 example 3.5b Factor: 16x2 + 40x + 25
Since a = +16 and c = +25 in the given trinomial, we are looking for two numbers that multiply to be (+16)(+25) = +400 and that also add to be +40. Since the 400 is positive, the two desired numbers will have the same sign…and, since the 40 is also positive, these signs will be positive. Therefore, the desired numbers are: +20, +20. Since both of these numbers are identical, the given trinomial is a perfect square. Thus, we can use a pattern to complete this task…
given polynomial: 16x2 +40x + 25 = (a + b)2 perfect square pattern: a2 +2ab + b2 = (a + b)2 Using the pattern: a2 = 16x2 = (4x)2
a = 4x
+2ab = +40x +2(4x)b = +40x
(+8x)b = +40x b = 5
b2 = 25 = (5)2 b = 5
Therefore: 16x2 + 40x + 25 = (4x + 5)2 = (4x + 5)(4x + 5)
Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more Click on the ad to read more
www.sylvania.com
We do not reinvent the wheel we reinvent light.
Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges.
An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.
Light is OSRAM
example 3.5c Factor: 4x2 – 9
Since a = 4 and c = –9 in the given trinomial, we are looking for two numbers that multiply to be (+4)(–9) = –36 and that also add to be zero (since b = 0).
Since the 36 is negative, the two desired numbers will have different signs…and, since b = 0, these two numbers will be opposites of each other. Therefore, the desired numbers are: ±6. Whenever these conditions exist, we have the difference of squares and we can use a pattern to complete this task…
given polynomial: 4x2 – 9 = (a + b)(a – b) difference of squares pattern: a2 – b2 = (a + b)(a – b) Using the pattern: a2 = 4x2 = (2x)2
a = 2x
b2 = 9 = (3)2 b = 3
Therefore: 4x2 – 9 = (2x + 3)(2x – 3) example 3.5d Factor: x3 – 12x2 + 48x – 64
Because this polynomial has four terms, we could attempt to group the terms to see if that strategy would work. However, we would soon discover that grouping is not the way to go for this example. Therefore, let’s attempt to see if the given polynomial fits one of our “special” patterns…
given polynomial: x3 – 12x2 + 48x – 64 = (a – b)3 perfect cube pattern: a3 – 3a2b + 3ab2 – b3 = (a – b)3 Using the pattern:
a3 = x3 a = x
3a2b = 12x2 3(x2)b = 12x2
3b = 12 b = 4
+3ab2 = +48x +3(x)b2 = +48x
+3b2 = +48 b2 = 16
b = 4
b3 = 64 = (4)3 b = 4
Therefore: x3 – 12x2 + 48x – 64 = (x – 4)3 = (x – 4)(x – 4)(x – 4) example 3.5e Factor: 8x3 + 36x2 + 54x + 27
Because this polynomial has four terms, we could attempt to group the terms to see if that strategy would work. However, we would soon discover that grouping is not the way to go for this example. Therefore, let’s attempt to see if the given polynomial fits one of our “special” patterns…
given polynomial: 8x3 + 36x2 + 54x + 27 = (a + b)3 perfect cube pattern: a3 + 3a2b + 3ab2 + b3 = (a + b)3
191 Using the pattern:
a3 = 8x3 = (2x)3 a = 2x
3a2b =36 3(2x)2b = 36x2 3(4x2)b = 36x2 (12x2)b = 36x2
b = 3
3ab2 = 54x 3(2x)b2 = 54x
(6x)b2 = 54x b2 = 9
b = 3
b3 = 27 = (3)3 b = 3
Therefore: 8x3 + 36x2 + 54x + 27 = (2x + 3)3 = (2x + 3)(2x + 3)(2x + 3) example 3.5f Factor: 64x3 + 27
Although this polynomial has two terms, it is not an example of the difference of two squares. Therefore, let’s attempt to see if the given polynomial fits one of our “special” patterns…
given polynomial: 64x3 + 27 = (a + b)(a2 – ab + b2) sum of cubes pattern: a3 + b3 = (a + b)(a2 – ab + b2) Using the pattern a3 = 64x3 = (4x)3→ a = 4x → a2 = (4x)2 = 16x2 b3 = 27 = (3)3→ b = 3 → b2 = (3)2 = 9
ab = (4x)(3) = 12x
Therefore: 64x3 + 27 = (4x + 3)(16x2 – 12x + 9) example 3.5g Factor: 125x3 – 64
Although this polynomial has two terms, it is not an example of the difference of two squares. Therefore, let’s attempt to see if the given polynomial fits one of our “special” patterns…
given polynomial: 125x3 – 64 = (a – b)(a2 + ab + b2) difference of cubes pattern: a3 – b3 = (a – b)(a2 + ab + b2) Using the pattern a3 = 125x3 = (5x)3→ a = 5x → a2 = (5x)2 = 25x2 b3 = 64 = (4)3→ b = 4 → b2 = (4)2 = 16
ab = (5x)(4) = 20x
Therefore: 125x3 – 64 = (5x – 4)(25x2 + 20x + 16)