A polynomial is an algebraic expression that contains one or more terms. The terms can only be added, subtracted or multiplied never divided. Multiplying polynomials is the process of combining all of the terms using the distributive property. Through this process, all of the like terms are combined. Multiplying polynomials combines all of the terms, leaving the polynomial in its extended form. Factoring polynomials is the opposite process, in which an extended form is broken down into its simplest forms.
How to Multiply Three Polynomials

Examine the expression (x + 6)(x – 9)(2x + 4^2). This statement is read: The quantity of x plus six times the quantity of x minus nine times the quantity of two x plus four squared. Rewrite the problem so that you are addressing the first two polynomials, (x + 6)(x – 9). It is easier to multiply in small manageable steps instead of trying to tackle the entire expression.

Multiply the two polynomials using the FOIL method. F stands for the first two terms, O stands for the outside terms, I is for the inside terms and L is for the last terms. Some people prefer to draw arrows to keep their multiplication steps organized.

Multiple the first two terms, X x X = x^2 or x squared. Multiply the outside terms, X x (9) = 9x. Multiply the inside terms, 6 x X= 6x. Multiply the last terms, 6 x (9) = 54. So far the multiplication answer should read x^2 – 9x + 6x – 54.

Combine like terms. You cannot combine the term x^2 with the other x’s because of the exponent, which can change the x factor drastically. Instead combine the single x’s, 9x + 6x = 3x. The answer so far reads x^2 – 3x – 54.

Rewrite the polynomial to include the remaining polynomial. (x^2 – 3x – 54)(2x + 4^2). First, do the math in the parenthesis, 4^2 = 16. Multiply with the FOIL process.

Multiply the first terms, x^2 by 2x = 2x^3. When multiplying exponents, multiply the base for a product of 2x and then add the exponents for the answer 2x^3. Multiply the outside terms, x^2 by 16= 16x^2. Multiply the inside term, 3x, by the first and last terms. 3x x 2x = 6x^2 and 3x x 16 = 48x. Multiply the last term, 54, by the first and last terms. 54 x 2x = 108x and 54 x 16 = 864. Your problem should read: 2x^3 + 16x^2 – 6x^2 – 48x – 108x – 864.

Combine like terms. 2x^3 + 10x^2 – 156x – 864.
How to Factor Three Polynomials

Examine the expression 5x^2 + 35x + 30.

Look for the Greatest Common Factor, in this case, 5 goes into all three terms. Write five outside of parenthesis, 5(…)(…). The inside is left blank for now, but will be filled in as the problem is factored.

Divide all three terms by the GFC, five. Factoring is the opposite of multiplying and gets expressions down to their simplest forms. Five goes into 5x^2 once, leaving only the x^2. Five goes into 35x seven times, leaving 7x. Five goes into 30, six times, leaving six. The answer so far is 5(x^2 + 7x + 6).

Factor out the parenthesis. First look at the first and last terms in the parenthesis. Are they squares, meaning can the numbers be broken down into a basic square root? No, they are not. X^2 is obviously squared but there is no square root of 6. Therefore, you’ll have to use trial and error to figure out the parenthesis’ simplest form.

Write a set of parenthesis, leaving the insides blank for now. Don’t forget the five from the first factoring step. 5(…)(…). Now, what do you need to make x a square? Another x. So fill this into the parenthesis. 5(x…)(x…). You can see that if you use FOIL, the first two terms equals x squared.

Factor out the terms for 6. They are 6 x 1 = 6 and 2 x 3 = 6. But which set of factors should you use so that the middle terms, the I in FOIL, add up to 7. The easiest way is to look at the factors. Does 2 + 3 equal 7? No, but 6 + 1 does. So write those factors into the parenthesis. 5(x…6)(x…1).

Select your sign. Because both 7x and 6 are positive, your signs will both be positive. 5(x + 6)(x + 1).

Multiply the parenthesis using FOIL to double check your work. X x X = x^2, X x 1 = x, 6 x X = 6x and 6 x 1 = 6. Combine like terms, X^2 + 7x + 6, which is the same as the problem after the second factoring step.
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