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Factoring Special Products
Overview
• Section 6.5 in the textbook
– Factoring perfect square trinomials
– Factoring the sum & difference of two squares
– Factoring the sum & difference of two cubes
– Factoring completely
Factoring Perfect Square
Trinomials
Notion of a Perfect Square
• A number n is a perfect square if we can
find an Integer k such that k · k = n
– The same Integer times itself
– Ex: 4 is a perfect square (k = 2)
81 is a perfect square (k = ?)
• A variable is a perfect square if its
exponent is evenly divisible by 2
– Ex:
p^{4} is a perfect square (4 is divisible by 2)
x^{3} is NOT a perfect square
Perfect Square Trinomials
• Remember to ALWAYS look for a GCF
before factoring!
• Consider what happens when we FOIL
(a + b)^{2}
(a + b)^{2} = a^{2} + 2ab + b^{2}
• a^{2} comes from squaring a in (a + b)^{2}
• 2ab comes from doubling the product of a
and b in (a + b)^{2}
• b^{2} comes from squaring b in (a + b)^{2}
Factoring Perfect Square
Trinomials
• To factor a perfect square trinomial, we
reverse the process:
– Answer the following questions:
• Are BOTH end terms a ^{2} and b^{2} perfect squares of
a and b respectively?
• Is the middle term two times a and b?
– If the answer to BOTH questions is YES, we can
factor a^{2} + 2ab + b^{2} as (a + b) (a + b) = (a + b)^{2}
– Otherwise, we must seek a new factoring
strategy
Ex 1: Factor completely:
a) x^{2}y^{2} – 8xy + 16y^{2}
b) 4r^{2} – 4r – 1
c) 36n^{2} + 36n + 9
Factoring the Sum &
Difference of Two Square
Difference of Two Squares
• Remember to ALWAYS look for a GCF
before factoring!
• 2 terms both of which are perfect squares
• Consider what happens when we FOIL
(a + b)(a – b)
(a + b)(a – b) = a^{2} – b^{2}
• a^{2}comes from the F term in (a + b)(a – b)
• b^{2}comes from the L term in (a + b)(a – b)
Factoring a Difference of Two
Squares
• To factor a difference of two squares, we
reverse the process:
– Answer the following questions:
• Are both terms a^{2} and b^{2} perfect squares of a and
b respectively?
• Is the sign between a^{2} and b^{2} a – ?
– If the answer to BOTH questions is YES,
a^{2} – b^{2} can be factored to (a + b)(a – b)
– Otherwise, the polynomial is prime
Ex 3: Factor completely:
a) x^{2} – 64y^{2}
b) 6z^{2} – 54
• Remember to ALWAYS look for a GCF before
factoring!
• Consider a^{2} + b^{2}
• Only 3 possibilities for the factoring:
(a + b)(a + b) = a^{2} + 2ab + b^{2} ≠ a^{2} + b^{2}
(a – b)(a – b) = a^{2} – 2ab + b^{2} ≠ a^{2} + b^{2}
(a + b)(a – b) = (a – b)(a + b) = a^{2} – b^{2} ≠ a^{2} + b^{2}
• Therefore, the sum of two squares is PRIME
Ex 4: Factor completely:
2x^{2} + 128
Factoring the Difference &
Sum of Two Cubes
Sum & Difference of Two Cubes
• Remember to ALWAYS look for a GCF
before factoring!
• Consider (a + b)(a^{2} – ab + b^{2})
a(a^{2} – ab + b^{2}) + b(a^{2} – ab + b^{2})
a^{3} – a^{2}b + ab^{2} + a^{2}b – ab^{2} + b^{3}
a^{3} + b^{3}
• Consider (a – b)(a^{2} + ab + b^{2})
a(a^{2} + ab + b^{2}) – b(a^{2} + ab + b^{2})
a^{3} – b^{3}
• Thus:
a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})
a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})
Factoring a Sum or Difference
of Two Cubes
• To factor a sum or difference of two cubes,
we reverse the process:
– Answer the following question:
• Are both terms a^{3} and b^{3} perfect cubes of a and b
respectively?
– If the answer is YES, a^{3} – b^{3} or a^{3} + b^{3} can be
factored into (a – b)(a^{2} + ab + b^{2}) or
(a + b)(a^{2} – ab + b^{2}) respectively
– Otherwise, the polynomial is prime
Ex 5: Factor completely:
a) x^{3} – 8
b) 27y^{3} + 64z^{3}
c) 250r^{3} – 2s^{3}
Factoring Completely
• Remember to ALWAYS look for a GCF
before factoring!
• Choose a factoring strategy based on the
number of terms
• Look at the result to see if any of the
products can be factored further
– Polynomials with a degree of 1 or less cannot
be factored further
• e.g. 2x + 1, 7
Ex 6: Factor completely:
a) x^{4} – 1
b) y^{4} – 16z^{4}
c) r^{4}t – s^{4}t
Summary
• After studying these slides, you should know
how to do the following:
– Recognize and factor a perfect square trinomial
– Factor a difference of two squares
– Recognize that the sum of two squares is prime
– Factor the difference or sum of two cubes
– Completely factor a polynomial
• Additional Practice
– See the list of suggested problems for 6.5
• Next lesson
– Solving Quadratic Equations by Factoring (Section
6.6)
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