Polynomial Review · Lesson 3
Three products come up so often that expanding them the long way wastes time and invites errors. Memorize the pattern once and you read the answer straight off. Two are perfect-square patterns; the third — difference of squares — is the single most useful identity in all of factoring, which you will run backwards constantly next lesson.
By the end, you can
(a + b)² = a² + 2ab + b². It is THREE terms: the square of the first, twice the product, the square of the last. Example: (x + 5)² = x² + 2·x·5 + 5² = x² + 10x + 25.
Why it works
(a+b)² = (a+b)(a+b). Distributing gives a² + ab + ba + b², and the two identical cross-terms fuse into 2ab. The middle term is the two ways the factors shake hands. Picture a square of side a+b: it splits into an a×a square, a b×b square, and TWO identical a×b rectangles.
(a − b)² = a² − 2ab + b². Same three-term shape; only the middle sign is negative, and the last term is +b² because a negative squared is positive. Example: (2x − 3)² = 4x² − 12x + 9.
Why it works
It is the sum pattern with b replaced by −b: then 2a(−b) = −2ab (middle turns negative) and (−b)² = +b² (last stays positive). The ending is always +, no matter the middle sign.
(a + b)(a − b) = a² − b². When you multiply a sum by its matching difference, the cross terms +ab and −ab cancel and the middle vanishes, leaving two terms. Example: (x + 7)(x − 7) = x² − 49.
Why it works
The four products are a², −ab, +ab, −b². The −ab and +ab are equal and opposite, so they annihilate. That 'middle cancels' behavior is the fingerprint you hunt for when factoring x² − 49 back into (x+7)(x−7).
Squaring a binomial is NOT squaring each term. The correct answer has a middle term: (x + 5)² = x² + 10x + 25. The exponent does not distribute over addition — (a+b)² is never a² + b².
Why it works
Test it with numbers: (3 + 4)² = 7² = 49, but 3² + 4² = 25. Not equal. The missing 2ab = 2·3·4 = 24 is exactly the gap: 25 + 24 = 49. Whenever you're tempted to square each piece, this 3-4-5 check reminds you the middle term is real.
Expand (x + 3)².
Apply the pattern
first² + 2·(first)·(last) + last² = x² + 2·x·3 + 3².
Expand (x − 4)².
Middle sign negative, last positive
x² − 2·x·4 + 4².
Expand (x + 9)(x − 9).
Recognize sum × difference
Same two pieces, opposite signs → a² − b² = x² − 9².
Expand (3x + 2)².
Square the first
(3x)² = 9x².
Twice the product
2·3x·2 = 12x.
Square the last
2² = 4.
Expand (5x − 4)(5x + 4).
Sum × difference of the same pieces
(5x)² − 4².
Expand (2x − 7)².
Square first, twice the product, square last
(2x)² − 2·(2x)·7 + 7² = 4x² − 28x + 49.
Note the ending
It is +49 even though the binomial had a minus.
Expand and simplify (x + 4)² − (x − 4)².
Expand both squares
(x² + 8x + 16) − (x² − 8x + 16).
Distribute the minus
x² + 8x + 16 − x² + 8x − 16.
Combine
The squares and constants cancel, leaving 16x.
(x + 5)² = x² + 25.
Missing the +10x. (x + 5)² = x² + 10x + 25. The 3-4-5 number test kills this instantly: (3+4)² = 49, not 3² + 4² = 25.
(x − 3)² = x² − 9.
That is the difference-of-squares answer, wrong product. The square of a difference is (x − 3)² = x² − 6x + 9.
(2x + 3)² = 2x² + …
Square the WHOLE first term: (2x)² = 4x², not 2x². So (2x + 3)² = 4x² + 12x + 9.
Making the last term of a difference-square negative: (x − 4)² = x² − 8x − 16.
A square is always +. (x − 4)² = x² − 8x + 16 — the last term is +16 because (−4)² = +16.
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t₁ = 5, d = 3 and t1=5,d=3 both count. For a list of numbers, separate them with commas (e.g. 3, 10, 17).Expand (x + 2)².
Expand (x − 6)².
Expand (x + 8)(x − 8).
Expand (x − 1)².
Is (x + 5)² equal to x² + 25?
Expand (x + 10)(x − 10).
Expand (4x + 1)².
Expand (3x − 5)².
Expand (6x + 7)(6x − 7).
Expand (2x − 7)².
(x − 3)² equals:
Expand and simplify (x + 3)² + (x − 3)².
Expand and simplify (x + 4)² − (x − 4)².
Using difference of squares, 102 · 98 =