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 Depdendent Variable

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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Simple Trinomials as Products of Binomials

Sometimes algebraic expressions of the form

ax 2 + bx + c (1)

where a, b, and c stand for actual numbers, can be written as a product of two binomials. This is considered a worthwhile property primarily when the original coefficients a, b, and c, and any numbers occurring in the resulting binomial factors are all whole numbers. When this situation occurs, it gives us a way of factoring such trinomials, rewriting them as a product of two somewhat simpler expressions.

Notice that

(x + a)(x + b) = x 2 + abx + ab

Thus, when we are given trinomial of the form

x 2 + dx + e

where the coefficient of x 2 is 1, we may be able to achieve this kind of factorization if we can find two whole numbers, a and b, such that

a + b = d,

the coefficient of x in the original trinomial, and,

ab = e,

the constant term in the trinomial.

(You might think that the problem is now quite simple, since we have two unknowns, a and b, and two equations, and so it’s just a matter of solving the system of equations. Unfortunately, there are two problems with this: most importantly, we are only interested in solutions which are whole number values, and also, the second equation here is not a linear equation. This doesn’t mean you can’t solve the system of equations given above in specific instances. In the example below, we will demonstrate a systematic inspection method which produces the values a and b required here when they actually exist.)

Example 1:

Factor x 2 + 5x + 6 if possible.

solution:

A minute’s examination indicates that these three terms contain no common monomial factors. However, since this expression has the pattern of expression (1) above with the coefficient of x 2 equal to 1, there is the possibility of factoring this expression into the product of two binomials, (x + a)(x + b). For this to be possible, we need to find two numbers, a and b, such that

a + b = 5

and

ab = 6.

We could just sit for a while and wait for inspiration, or we could test out a few guesses. However, we can be a bit more systematic. The more restrictive condition is that ab = 6. There aren’t many possibilities. We’ll make a little table listing all of the pairs of whole numbers whose product is 6:

 a b a + b 1 6 7 2 3 5 -1 -6 -7 -2 -3 -5

(You might think that a = 6, b = 1, for instance, should also be in our table. However, swapping values of a and b just amounts to reversing the order of the factors in the final product, and so does not give any new possibilities. It is important to consider both positive and negative factors which multiply to give +6 in this case, however.)

From this table, we see that there are only four pairs of whole numbers which multiply to give +6. We’ve also listed the sum of the two numbers in each case, and you can see that one of these pairs of values does sum to +5. Thus, a = 2 and b = 3 seems to satisfy the requirements here. Checking,

(x + 2) (x + 3) = (x + 2)(x) + (x + 2)(3)

= x(x + 2) + 3(x + 2)

= x 2 + 2x + 3x + 6

= x 2 + 5x + 6

which is identical to the original expression in this example. Hence, in factored form we can write

x 2 + 5x + 6 = (x + 2) (x + 3)

(Note that if the table above containing pairs of whole numbers with a product of +6 had contained no row in which the two numbers summed to +5, this would be proof that the type of factorization we were attempting could not be achieved. Thus, this systematic inspection method provides a factorization of a trinomial like this when such a factorization exists, and demonstrates that no such factorization is possible when that is the case.)