Starting with a polynomial in standard form, you will study how to change a polynomial equation into a product of its individual terms.
We will study the four methods of factoring, in addition to the preferred “trial and error” method:
Polynomials can be factored by applying the distributive property by pulling out the common term of the polynomial. First find the greatest common factor (GCF).
For example, first find the GCF of .
Each term can be written as a product of individual terms:
Remove the GCF from each term. Factoring out the GCF from each term the polynomial is factored as .
We can verify our answer by distributing the 7x across both terms in parentheses to reproduce our original equation .
Which choice shows the correct factorization of by factoring out the GCF?
The correct answer is C. To find the GCF, expand each term.
From this, we can see that the GCF is 2x, which can be factored out of each term in the original polynomial.
Multiplying two binomials using the FOIL method results in a polynomial in standard form.
To make the process easier, group the coefficients together as for the linear term and rt for the constant term.
Comparing this to the standard form, we can see that the coefficient b on the linear term is equal to . The constant c is equal to rt. We now have two equations with two unknowns. Now find two numbers that, when added, result in b and when multiplied, result in c.
Trial and error becomes important in this step. Try different numbers, both positive and negative, that will satisfy these requirements.
Which are the factors of the trinomial ?
The correct answer is D. In this equation, r and t must add to equal 5, and they must multiply to equal 6.
By trial and error (and inspection), or solving for each variable, we can conclude that r and t equal 3 and 2. Then, simply write the factors .
If we use FOIL and multiply the two binomials (assuming the factors are correct), we will reproduce the original trinomial . If not, there was an error in the selection of r and t, and the process must be repeated until the correct numbers are found.
It is very important to take a few extra minutes to check that you have selected the correct factors. Factoring more complex trinomials can be difficult, and using FOIL to check your answer is an important step that must not be overlooked.
Now there is a coefficient on the quadratic term equal to a and a third equation must be added to help find its value. With the addition of this coefficient, factoring the quadratic equation will produce two binomials in the form .
The equations become ,
, and
. From here, the goal is to find m, n, r, and t. Remember that, as long as the trinomial is not too difficult, you can probably use trial and error to find the solution.
Which is the correct factorization of ?
The correct answer is A. The following information is known:
We see that m = 2, n = 1, and r = 1 and t = 5. In factored form,
.
Sometimes polynomials that initially seem complex can be factored easily by noticing special cases.
Difference of two squares :
Perfect Square Trinomials :
Which shows the correct factorization of ?
The correct answer is B. Notice that the a-term and the b-term are perfect squares. We can write this trinomial as , which fits the perfect square trinomial formula. In factored form, this is
.
Expand the binomial to check the answer.
Which choice shows the correct factorization of ?
The correct answer is D. This is a difference of squares polynomial and can be written in the form . Apply the formula.
Use the FOIL method to check accuracy.
Factoring is one method that lets us find the solution, or roots, of the polynomial. In graphical terms, this is the point or points where the graph crosses the x-axis and where the polynomial is equal to zero.
For example, three roots of a polynomial function are –2, 3, and 3. We can then write the following function in factored form:
Notice that is equal to 0 if –2, 3, or 3 is substituted into this equation. Use the FOIL method to combine the first two binomials.
Use the distributive property to combine this trinomial and the third binomial.
If we draw the graph of , it is evident what roots mean.
On the x–y coordinate plane, the graph touches and reflects off of the x-axis at the point x = 3, and crosses the x-axis at x = -2. Because the graph reflects and does not cross the axis at x = 3, we have two identical roots at this point. Therefore, the roots for this polynomial are located at –2, 3, and 3.