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Expand special products of binomials

Understanding includes deriving and applying the identities (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2, (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2, and (a+b)(a−b)=a2−b2(a+b)(a-b)=a^2-b^2 for algebraic expressions with integer or rational coefficients. The learner connects these forms to the distributive property, recognizing the doubled middle term in a binomial square and the cancellation of cross terms in conjugate products; the scope is limited to these standard quadratic identities, not higher-degree or abstract generalizations.

Detailed Explanation: Expand special products of binomials

Special products are shortcuts based on the distributive property. The three important identities are

(a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2

and

(a+b)(a−b)=a2−b2.(a+b)(a-b)=a^2-b^2.

The middle term in a squared binomial is doubled because the two cross-products are alike. In conjugate products, the cross-products cancel.

Example: Expand ((2x−3)2)((2x-3)^2)

Recognize this as the pattern

(a−b)2=a2−2ab+b2.(a-b)^2=a^2-2ab+b^2.

Here,

a=2xandb=3.a=2x \qquad \text{and} \qquad b=3.

Substitute these values into the identity:

(2x−3)2=(2x)2−2(2x)(3)+32.(2x-3)^2=(2x)^2-2(2x)(3)+3^2.

Now simplify each term:

(2x)2=4x2,(2x)^2=4x^2, −2(2x)(3)=−12x,-2(2x)(3)=-12x,

and

32=9.3^2=9.

Therefore,

(2x−3)2=4x2−12x+9.\boxed{(2x-3)^2=4x^2-12x+9}.

Be careful not to forget the middle term. For a squared binomial, it is always twice the product of the two terms, with the sign determined by the original binomial.

Learn by doing: Expand special products of binomials

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