How do you factor the perfect square trinomial?

How do you factor the sum and difference of two cubes?

Which of these three makes the most sense to you? Explain why.

How do you factor the perfect square trinomial?

How do you factor the sum and difference of two cubes?

Which of these three makes the most sense to you? Explain why.

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Every perfect square trinomial has the form

x^{2n} + 2ax^{n} + a^{2}.

where n is a positive integer.

It can factor as (x^{n} + a)^{2}

Difference of two n^{th} powers:

n = 2 → a^{2} - b^{2} = (a - b)(a + b) Squares.

n = 3 → a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) Cubes.

n = 4 → a^{4} - b^{4} = (a - b)(a^{3} + a^{2}b + ab^{2} + b^{3}).

n = 5 → a^{5} - b^{5} = (a - b)(a^{4} + a^{3}b + a^{2}b^{2} + ab^{3} + b^{4}).

Can you see the pattern?

For odd n you can also factor sums of two n^{th} powers

n = 3 → a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2}) Cubes.

n = 5 → a^{5} + b^{5} = (a + b)(a^{4} - a^{3}b + a^{2}b^{2} - ab^{3} + b^{4}).

n = 7 → a^{7} + b^{7} = (a + b)(a^{6} - a^{5}b + a^{4}b^{2} - a^{3}b^{3} + a^{2}b^{4} - ab^{5} +b^{6}).

Can you see the pattern?

Also, for both sums and differences of two n^{th} powers:

1. If n is prime, the second factor is a prime polynomial and so no further factorization can be done.

2. If n is composite, the second factor can be factored using factoring by grouping.

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