Notable Products in Polynomial Multiplication
Notable products are special cases in polynomial multiplication that allow you to find the result directly - without expanding all the intermediate steps.
Sum and difference of two terms
$$ (a+b) \cdot (a-b) = a^2 - b^2 $$
Square of a binomial
$$ (a+b)^2 = a^2 + b^2 + 2ab $$ $$ (a-b)^2 = a^2 + b^2 - 2ab $$
Cube of a binomial
$$ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $$ $$ (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 $$
Difference of squares
$$ a^2 - b^2 = (a+b) \cdot (a-b) $$
Sum and difference of cubes
$$ x^3-a^3 = (x-a) \cdot (x^2+ax+a^2) $$ $$ x^3+a^3 = (x+a) \cdot (x^2-ax+a^2) $$
Square of a trinomial
$$ (a+b+c)^2 = a^2 + b^2 +c^2 + 2ab +2ac + 2bc $$
Quadratic trinomial (special case)
$$ x^2+(a+b)x+ ab = (x+a)(x+b) $$
Product of the Sum and Difference of Two Terms
Multiplying the sum and difference of the same two terms yields the difference of their squares.
$$ (a+b) \cdot (a-b) = a^2 - b^2 $$
Example. Let’s take two polynomials formed by the sum and difference of the same terms: $$ P(x): \ 4a + 2b $$ $$ Q(x): \ 4a - 2b $$ We apply the special product formula: $$ P(x) \cdot Q(x) = (4a+2b)(4a-2b) $$ $$ P(x) \cdot Q(x) = (4a)^2 - (2b)^2 $$ So the result is: $$ P(x) \cdot Q(x) = 16a^2 - 4b^2 $$
Verification. To confirm, let’s expand the product step by step: $$ P(x) \cdot Q(x) = (4a+2b)(4a-2b) $$ $$ = 4a(4a - 2b) + 2b(4a - 2b) $$ $$ = 16a^2 - 8ab + 8ab - 4b^2 $$ $$ = 16a^2 - 4b^2 $$ The final result matches our earlier computation.
Square of a Binomial
The square of a binomial (a+b)2 equals the sum of the squares of each term plus twice their product.
$$ (a+b)^2 = a^2 + b^2 + 2ab $$
Example. Let’s compute the square of a binomial: $$ (2a+3b)^2 $$ Expanding it using the formula: $$ (2a+3b)^2 = (2a)^2 + (3b)^2 + 2 \cdot (2a \cdot 3b) $$ $$ = 4a^2 + 9b^2 + 12ab $$
Cube of a Binomial
The cube of a binomial (a+b)3 is equal to the cube of the first term, plus three times the square of the first times the second, plus three times the first times the square of the second, plus the cube of the second.
$$ (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 $$
Example. Let’s expand the cube of a binomial: $$ (2a-3b)^3 $$ Using the formula: $$ (2a-3b)^3 = (2a)^3 + (3b)^3 + 3 \cdot (2a)^2 \cdot (-3b) + 3 \cdot 2a \cdot (-3b)^2 $$ $$ = 8a^3 + 27b^3 - 36a^2b + 54ab^2 $$
Sum and Difference of Cubes
The sum or difference of two cubes can be factored into a binomial multiplied by a trinomial.
$$ x^3-a^3 = (x-a)(x^2 + ax + a^2) $$ $$ x^3+a^3 = (x+a)(x^2 - ax + a^2) $$
Example. Consider the expression: $$ x^3 - 27 $$ This is a difference of cubes: $$ x^3 - 3^3 $$ Applying the identity: $$ (x - 3)(x^2 + 3x + 9) $$
Quadratic Trinomial in the Form x2 + sx + p
A second-degree trinomial in standard form looks like: x² + sx + p
$$ x^2 + s \cdot x + p $$
Here, the leading coefficient is 1, the coefficient of x is the sum of two numbers s = a + b, and the constant term is their product p = ab.
$$ x^2 + (a+b)x + ab $$
This can be factored as: (x + a)(x + b)
$$ x^2 + (a+b)x + ab = (x+a)(x+b) $$
Example. Let’s factor the trinomial: $$ x^2 + 5x + 6 $$ The coefficient of x is the sum 2 + 3, and the constant term is their product: 2·3 = 6. $$ x^2 + (2 + 3)x + 2 \cdot 3 $$ This matches the identity, so we factor it as: $$ x^2 + 5x + 6 = (x+2)(x+3) $$
Verification. Expanding the product to check: $$ (x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6 $$ The factorization is correct.
And so on.
