Remarkable products. Frequently, multiplications are presented in algebra whose products meet fixed rules. These products are called notable products.
Summary
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- 1 Square of the sum of two terms
- 2 Square difference of two terms
- 3 Sum by the difference of two terms
- 4 Factorial decomposition. Common factor extraction
- 5 Common factor
- 6 Factorial decomposition of binomials
- 7 Factorial decomposition of trinomials
- 8 Trinomial of the form x2 + px + q
- 9 Trinomial of the form mx2 + px + q, m ≠ 1
- 10 Sources
Square of the sum of two terms
Squared a + b is equivalent to multiplying this binomial by itself, then it turns out: (a + b) 2 = (a + b) (a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 OR let: (a + b) 2 = a2 + 2ab + b2 That is, the square of a sum is equal to the square of the first term, plus the double of the first term by the second, plus the square of the second term.
Square of the difference of two terms
(ab) 2 = (ab) (ab) = a2-ab-ab + b2 = a2-2ab + b2 In other words: (ab) 2 = a2-2ab + b2 We then have the square of a difference equal to square of the first term, minus the double of the first term by the second, plus the square of the second. These notable products differ only in their sign, so we can summarize them as follows (a ± b) 2 = a2 ± 2ab + b2
Sum by the difference of two terms
(a + b) (ab) = a2-ab + ab-b2 = a2-b2 Then (a + b) (ab) = a2-b2 Therefore the product of the sum of the difference of two terms is equal unlike their squares. Product of two binomials that have a common term. (x + a) (x + b) = x2 + bx + ax + ab = x2 + (a + b) x + ab bx + ax = (a + b) x Then (x + a) (x + b) = x2 + (a + b) x + ab Therefore the product of two binomials that have a common term is equal to the square of the common term, plus the algebraic sum of the uncommon terms by the common term, plus the product of the non-common terms. common.
Factorial decomposition. Common factor extraction
If two algebraic expressions A and B are multiplied and their product is C, each of the expressions A and B is said to be a factor or divisor of C. Example, since a (a + b) = a2 + ab, we says that a and a + b are factors or divisors of a2 + ab. Similarly (x + 2) (x -2) = x2 -4, then x + 2 and x-2 are factors or divisors of x2 -4. It is often convenient to determine the factors of a given algebraic expression C. The operation of finding these factors (when they exist) is called factorization or factor decomposition of expression C.
Common factor
To factor the expression ma + mb, apply the distributive property: ma + mb = m (a + b) where m is the common factor of the expression ma + mb, since this factor appears in each of the terms of the given expression. In general, if in a given algebraic expression there is a factor that is common to all its terms, it can be decomposed into the product of that common factor by the polynomial that results from dividing each of the terms of the given expression by that common factor. .
Factorial decomposition of binomials
The sum of two terms multiplied by their difference equals the difference of their squares: (a + b) (ab) = a2-b2 Using this remarkable product, a procedure can be obtained to factorize a difference of two squares The difference of two squares is decomposed into the product of the sum by the difference of the bases of their squares In symbols: a2-b2 = (a + b) (ab)
Factorial decomposition of trinomials
Perfect square trinomial . Any trinomial that is a perfect square can be transformed into the square of a binomial. A trinomial is a perfect square if: • Two of its terms are perfect squares, and • The remaining term is equal to the double product of the square roots of said terms, or the opposite of that product.
The trinomial is decomposed into the square of the sum or difference of the square roots of the perfect square terms, depending on whether the sign of the remaining term is positive or negative. a2 ± 2ab + b2 = (a ± b) 2
Trinomial of the form x2 + px + q
A trinomial of the form x2 + px + q can be decomposed into the product of two factors (x + a) and (x + b) as long as we can find two numbers a and b whose algebraic sum is p and whose product is q, and then we have : x2 + px + q = (x + a) (x + b), where a + b = p and ab = q.
Trinomial of the form mx2 + px + q, m ≠ 1
In general, (ax + b) (cx + d) = acx2 + (ad + bc) x + bd = m x2 + px + q Where m = ac, q = bd and p = ad + bc. Therefore, whenever possible to determine the numbersa, b, c and d, such that: ac = m, bd = q and ad = bc = p is true m x2 + px + q = (ax + b) (cx + d) The trinomials of the form mx2 + px + q, ( m ≠ 1) can also be factorized by reducing them to the form x2 + px + q. For this, the following procedure is used: The given trinomial is multiplied and divided by m, with which we obtain: m x2 + px + q = (mx) 2 + p (mx) + mq m Thus, the numerator is reduced to the form x2 + px + q and considering mx as a single variable, this numerator is decomposed into a product of the form (mx + a) (mx + b) whenever it is possible to find two numbers that multiply give mq and that added together give p.
