5.3: Multiplying Polynomials
- Page ID
- 116786
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- Multiply a polynomial by a binomial.
- Multiply a polynomial by any size polynomial.
- Recognize and calculate special products.
- Multiply polynomial functions.
Multiplying by a Monomial
Recall the product rule for exponents: if \(m\) and \(n\) are positive integers, then
\[x^{m}\cdot x^{n}=x^{m+n}\]
In other words, when multiplying two expressions with the same base, add the exponents. This rule applies when multiplying a monomial by a monomial. To find the product of monomials, multiply the coefficients and add the exponents of variable factors with the same base. For example,
\(\begin{array} {cl} {3x\cdot 5x^{2} = 3\cdot 5\cdot x^{1}\cdot x^{2}}&{\color{Cerulean}{Commutative\: property}}\\{=15x^{1+2}}&{\color{Cerulean}{Product\:rule\:for\:exponents}}\\{=15x^{3}}&{} \end{array}\)
To multiply a polynomial by a monomial, apply the distributive property and then simplify each term.
Multiply:
\(−5x(4x−2)\).
Solution:
In this case, multiply the monomial, \(−5x\), by the binomial, \(4x−2\). Apply the distributive property and then simplify.
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Answer:
\(-20x^{2}+10x\)
Multiply:
\(2x^{2}(3x^{2}−5x+1)\).
Solution:
Apply the distributive property and then simplify.
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Answer:
\(6x^{4}-10x^{3}+2x^{2}\)
Multiply:
\(−3ab^{2}(a^{2}b^{3}+2a^{3}b−6ab−4)\).
Solution:
Answer:
To summarize, multiplying a polynomial by a monomial involves the distributive property and the product rule for exponents. Multiply all of the terms of the polynomial by the monomial. For each term, multiply the coefficients and add exponents of variables where the bases are the same.
Multiply:
\(−5x^{2}y(2xy^{2}−3xy+6x^{2}y−1)\).
- Answer
-
\(−10x^{3}y^{3}+15x^{3}y^{2}−30x^{4}y^{2}+5x^{2}y\)
Multiplying by a Binomial
In the same way that we used the distributive property to find the product of a monomial and a binomial, we will use it to find the product of two binomials.
\[\begin{aligned} \color{Cerulean}{(a+b)}\color{black}{(c+d)} \\ &=\color{Cerulean}{(a+b)}\color{black}{\cdot c+}\color{Cerulean}{(a+b)}\color{black}{\cdot d} \\ &=ac+bc+ad+bd \\ &=ac+ad+bc+bd \end{aligned}\]
Here we apply the distributive property multiple times to produce the final result. This same result is obtained in one step if we apply the distributive property to \(a\) and \(b\) separately as follows:
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This is often called the FOIL method. We add the products of the first terms of each binomial \(ac\), the \(o\)uter terms \(ad\), the \(i\)nner terms \(bc\), and finally the last terms \(bd\). This mnemonic device only works for products of binomials; hence it is best to just remember that the distributive property applies.
Multiply:
\((2x+3)(5x−2)\).
Solution:
Distribute \(2x\) and then distribute \(3\).
Simplify by combining like terms.
\(=10x^{2}+11x-6\)
Answer:
\(10x^{2}+11x-6\)
Multiply:
\((\frac{1}{2}x−\frac{1}{4})(\frac{1}{2}x+\frac{1}{4})\).
Solution:
Distribute \(\frac{1}{2}x\) and then distribute \(−\frac{1}{4}\).
\(\begin{aligned} (\frac{1}{2}x−\frac{1}{4})(\frac{1}{2}x+\frac{1}{4}) &=\color{Cerulean}{\frac{1}{2}x}\color{black}{\frac{1}{2}x+}\color{Cerulean}{\frac{1}{2}x}\color{black}{\cdot\frac{1}{4}+}\color{OliveGreen}{\left( -\frac{1}{4} \right)}\color{black}{\cdot\frac{1}{2}x+}\color{OliveGreen}{\left(-\frac{1}{4} \right)}\color{black}{\cdot\frac{1}{4}} \\ &=\frac{1}{4}x^{2}+\frac{1}{8}x-\frac{1}{8}x-\frac{1}{16} \\ &=\frac{1}{4}x^{2}-\frac{1}{16} \end{aligned}\)
Answer:
\(\frac{1}{4}x^{2}-\frac{1}{16}\)
Multiply:
\((3y^{2}−1)(2y+1)\).
Solution:
Answer:
\(6y^{3}+3y^{2}-2y-1\)
After applying the distributive property, combine any like terms.
Multiply:
\((x^{2}−5)(3x^{2}−2x+2)\).
Solution:
After multiplying each term of the trinomial by \(x^{2}\) and \(−5\), simplify.
Answer:
\(3x^{4}-2x^{3}-13x^{2}+10x-10\)
Multiply:
\((2x−1)^{3}\).
Solution:
Perform one product at a time.
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Answer:
\(8x^{3}-12x^{2}+6x-1\)
At this point, it is worth pointing out a common mistake:
\((2x-1)^{3}\neq (2x)^{3}-(1)^{3}\)
The confusion comes from the product to a power rule of exponents, where we apply the power to all factors. Since there are two terms within the parentheses, that rule does not apply. Care should be taken to understand what is different in the following two examples:
\(\begin{aligned} (xy)^{2} &=x^{2}y^{2}\quad\color{Cerulean}{\checkmark} \\ (x+y)^{2} &\neq x^{2}+y^{2}\quad\color{red}{x} \end{aligned}\)
Multiply:
\((2x−3)(7x^{2}−5x+4)\).
- Answer
-
\(14x^{3}-31x^{2}+23x-12\)
Product of Polynomials
When multiplying polynomials, we apply the distributive property many times. Multiply all of the terms of each polynomial and then combine like terms.
Multiply:
\((2x^{2}+x−3)(x^{2}−2x+5)\).
Solution:
Multiply each term of the first trinomial by each term of the second trinomial and then combine like terms.
Aligning like terms in columns, as we have here, aids in the simplification process
Answer:
\(2x^{4}-3x^{3}+5x^{2}+11x-15\)
Notice that when multiplying a trinomial by a trinomial, we obtain nine terms before simplifying. In fact, when multiplying an \(n\)-term polynomial by an m-term polynomial, we will obtain \(n × m\) terms. In the previous example, we were asked to multiply and found that
\((2x^{2}+x-3)(x^{2}-2x+5)=2x^{4}-3x^{3}+5x^{2}+11x-15\)
Because it is easy to make a small calculation error, it is a good practice to trace through the steps mentally to verify that the operations were performed correctly. Alternatively, we can check by evaluating any value for \(x\) in both expressions to verify that the results are the same. Here we choose \(x = 2\):
\(\begin{aligned} (2x^{2}+x-3)(x^{2}-2x+5)&=(2(\color{OliveGreen}{2}\color{black}{)^{2}+(}\color{OliveGreen}{2}\color{black}{)-3)((}\color{OliveGreen}{2}\color{black}{)^{2}-2(}\color{OliveGreen}{2}\color{black}{)+5)} \\ &=(8+2-3)(4-4+5) \\ &=(7)(5) \\ &=35 \end{aligned}\)
Because the results could coincidentally be the same, a check by evaluating does not necessarily prove that we have multiplied correctly. However, after verifying a few values, we can be fairly confident that the product is correct.
Exercise \(\PageIndex{3}\)
Multiply:
\((x^{2}−2x−3)^{2}\).
- Answer
-
\(x^{4}−4x^{3}−2x^{2}+12x+9\)
Special Products
In this section, the goal is to recognize certain special products that occur often in our study of algebra. We will develop three formulas that will be very useful as we move along. The three should be memorized. We begin by considering the following two calculations:
\(\begin{array}{r|r} {(a+b)^{2}=(a+b)(a+b)}&{(a-b)^{2}=(a-b)(a-b)}\\{=a^{2}+ab+ba+b^{2}}&{=a^{2}-ab-ba+b^{2}}\\{=a^{2}+ab+ab+b^{2}}&{=a^{2}-ab-ab+b^{2}}\\{=a^{2}+2ab+b^{2}}&{=a^{2}-2ab+b^{2}} \end{array}\)
This leads us to two formulas that describe perfect square trinomials:
\[ (a+b)^{2}=a^{2}+2ab+b^{2} \]
\[ (a-b)^{2}=a^{2}-2ab+b^{2}\]
We can use these formulas to quickly square a binomial.
Multiply:
\((3x+5)^{2}\).
Solution:
Here \(a=3x\) and \(b=5\). Apply the formula:
\(\begin{aligned} \color{Cerulean}{(a+b)^{2}} &\color{Cerulean}{ =\:\: a^{2}\:\:\:\:+2\:\:\:\:\:\:a\:\:\:\:\:b\:\:+\:\:b^{2}} \\ &\color{Cerulean}{\quad\:\:\: \downarrow\qquad\qquad\:\: \downarrow\:\:\:\:\: \downarrow\qquad\downarrow} \\ (3x+5)^{2}&=(3x)^{2}+2\cdot(3x)(5)+(5)^{2} \\ &=9x^{2}+30x+25\end{aligned}\)
Answer:
\(9x^{2}+30x+25\)
This process should become routine enough to be performed mentally.
Multiply:
\((x−4)^{2}\).
Solution:
Here \(a=x\) and \(b=4\). Apply the appropriate formula as follows:
\(\begin{aligned} \color{Cerulean}{(a-b)^{2}} &\color{Cerulean}{ =\:\: a^{2}\:\:\:\:-2\:\:\:\:a\:\:\:b\:\:+\:b^{2}} \\ &\color{Cerulean}{\quad\:\:\: \downarrow\qquad\quad\:\:\:\: \downarrow\:\:\: \downarrow\quad\:\:\:\downarrow} \\ (x-4)^{2}&=(x)^{2}-2\cdot(x)(4)+(4)^{2} \\ &=x^{2}-8x+16\end{aligned}\)
Answer:
\(x^{2}-8x+16\)
Our third special product follows:
\(\begin{aligned}(a+b)(a-b)&=a^{2}-ab+ba-b^{2} \\ &=a^{2}\color{red}{-ab+ab}\color{black}{-b^{2}}\\&=a^{2}-b^{2} \end{aligned}\)
This product is called difference of squares:
\[(a+b)(a-b)=a^{2}-b^{2}\]
The binomials \((a+b)\) and \((a−b)\) are called conjugate binomials. Therefore, when conjugate binomials are multiplied, the middle term eliminates, and the product is itself a binomial.
Multiply:
\((7x+4)(7x−4)\).
Solution:
Answer:
\(49x^{2}-16\)
Multiply:
\((−5x+2)^{2}\).
- Answer
-
\(25x^{2}−20x+4\)
Multiplying Polynomial Functions
We use function notation to indicate multiplication as follows:
| Multiplication of functions: | \((f\cdot g)(x)=f(x)\cdot g(x)\) |
|---|
Calculate:
\((f⋅g)(x)\), given \(f(x)=5x^{2}\) and \(g(x)=−x^{2}+2x−3\).
Solution:
Multiply all terms of the trinomial by the monomial function \(f(x)\).
\(\begin{aligned} (f\cdot g)(x)&=f(x)\cdot g(x) \\ &=5x^{2}\cdot (-x^{2}+2x-3) \\ &=-5x^{4}+10x^{3}-15x^{2} \end{aligned}\)
Answer:
\((f\cdot g)(x)=-5x^{4}+10x^{3}-15x^{2}\)
Calculate:
\((f⋅g)(−1)\), given \(f(x)=−x+3\) and \(g(x)=4x^{2}−3x+6\).
Solution:
First, determine \((f⋅g)(x)\).
\(\begin{aligned} (f\cdot g)(x) &=f(x)\cdot g(x) \\ &=(-x+3)(4x^{2}-3x+6) \\ &=-4x^{3}+3x^{2}-6x+12x^{2}-9x+18 \\ &=-4x^{3}+15x^{2}-15x+18 \end{aligned}\)
We have
\((f\cdot g)(x) = -4x^{3}+15x^{2}-15x+18
Next, substitute \(−1\) for the variable \(x\).
\(\begin{aligned} (f\cdot g)(\color{OliveGreen}{-1}\color{black}{)}&=-4(\color{OliveGreen}{-1}\color{black}{)^{3}+15(}\color{OliveGreen}{-1}\color{black}{)^{2}-15(}\color{OliveGreen}{-1}\color{black}{)+18} \\ &=-4\cdot (-1)+15\cdot 1+15+18 \\ &=4+15+15+18 \\ &=52 \end{aligned}\)
Answer:
\((f\cdot g)(-1)=52\)
Because \((f⋅g)(−1)=f(−1)⋅g(−1)\), we could alternatively calculate \(f(−1)\) and \(g(−1)\) separately and then multiply the results (try this as an exercise). However, if we were asked to evaluate multiple values for the function \((f⋅g)(x)\), it would be best to first determine the general form, as we have in the previous example.
Key Takeaways
- To multiply a polynomial by a monomial, apply the distributive property and then simplify each of the resulting terms.
- To multiply polynomials, multiply each term in the first polynomial with each term in the second polynomial. Then combine like terms.
- The product of an \(n\)-term polynomial and an \(m\)-term polynomial results in an \(m × n\) term polynomial before like terms are combined.
- Check results by evaluating values in the original expression and in your answer to verify that the results are the same.
- Use the formulas for special products to quickly multiply binomials that occur often in algebra.
Multiply.
- \(5x(−3x^{2}y)\)
- \((−2x^{3}y^{2})(−3xy^{4})\)
- \(\frac{1}{2}(4x−3)\)
- \(−\frac{3}{4}(\frac{2}{3}x−6)\)
- \(3x(5x−2)\)
- \(−4x(2x−1)\)
- \(x^{2}(3x+2)\)
- \(−6x^{2}(5x+3)\)
- \(2ab(4a−2b)\)
- \(5a^{2}b(a^{2}−b^{2})\)
- \(6x^{2}y^{3}(−3x^{3}y+xy^{2})\)
- \(3ab^{3}(−5ab^{3}+6a^{2}b)\)
- \(−\frac{1}{2}x^{2}y(4xy−10)\)
- \(−3x^{4}y^{2}(3x^{8}y^{3})\)
- \(2x^{2}(−5x^{3})(3x^{4})\)
- \(4ab(a^{2}b^{3}c)(a^{4}b^{2}c^{4})\)
- \(−2(5x^{2}−3x+4)\)
- \(\frac{4}{5}(25x^{2}−50xy+5y^{2})\)
- \(3x(5x^{2}−2x+3)\)
- \(−x(x^{2}+x−1)\)
- \(x^{2}(3x^{2}−5x−7)\)
- \(x^{3}(−4x^{2}−7x+9)\)
- \(\frac{1}{4}x^{4}(8x^{3}−2x^{2}+\frac{1}{2}x−5)\)
- \(−\frac{1}{3}x^{3}(\frac{3}{2}x^{5}−\frac{2}{3}x^{3}+\frac{9}{2}x−1)\)
- \(a^{2}b(a^{2}−3ab+b^{2})\)
- \(6a^{2}bc^{3}(2a−3b+c^{2})\)
- \(\frac{2}{3}xy^{2}(9x^{3}y−27xy+3xy^{3})\)
- \(−3x^{2}y^{2}(12x^{2}−10xy−6y^{2})\)
- Find the product of \(3x\) and \(2x^{2}−3x+5\).
- Find the product of \(−8y\) and \(y^{2}−2y+12\).
- Find the product of \(−4x\) and \(x^{4}−3x^{3}+2x^{2}−7x+8\).
- Find the product of \(3xy^{2}\) and \(−2x^{2}y+4xy−xy^{2}\).
- Answer
-
1. \(−15x^{3}y\)
3. \(2x−\frac{3}{2}\)
5. \(15x^{2}−6x\)
7. \(3x^{3}+2x^{2}\)
9. \(8a^{2}b−4ab^{2}\)
11. \(−18x^{5}y^{4}+6x^{3}y^{5}\)
13. \(−2x^{3}y^{2}+5x^{2}y\)
15. \(−30x^{9}\)
17. \(−10x^{2}+6x−8\)
19. \(15x^{3}−6x^{2}+9x\)
21. \(3x^{4}−5x^{3}−7x^{2}\)
23. \(2x^{7}−\frac{1}{2}x^{6}+\frac{1}{8}x^{5}−\frac{5}{4}x^{4}\)
25. \(a^{4}b−3a^{3}b^{2}+a^{2}b^{3}\)
27. \(6x^{4}y^{3}−18x^{2}y^{3}+2x^{2}y^{5}\)
29. \(6x^{3}−9x^{2}+15x\)
31. \(−4x^{5}+12x^{4}−8x^{3}+28x^{2}−32x\)
Multiply.
- \((3x−2)(x+4) \)
- \((x+2)(x−3) \)
- \((x−1)(x+1) \)
- \((3x−1)(3x+1) \)
- \((2x−5)(x+3) \)
- \((5x−2)(3x+4) \)
- \((−3x+1)(x−1) \)
- \((x+5)(−x+1) \)
- \((y−\frac{2}{3})(y+\frac{2}{3})\)
- \((\frac{1}{2}x+\frac{1}{3})(\frac{3}{2}x−\frac{2}{3})\)
- \((\frac{3}{4}x+\frac{1}{5})(\frac{1}{4}x+\frac{2}{5})\)
- \((\frac{1}{5}x+\frac{3}{10})(\frac{3}{5}x−\frac{5}{2})\)
- \((y^{2}−2)(y+2)\)
- \((y^{3}−1)(y^{2}+2)\)
- \((a^{2}−b^{2})(a^{2}+b^{2})\)
- \((a^{2}−3b)^{2}\)
- \((x−5)(2x^{2}+3x+4) \)
- \((3x−1)(x^{2}−4x+7) \)
- \((2x−3)(4x^{2}+6x+9) \)
- \((5x+1)(25x^{2}−5x+1) \)
- \((x−\frac{1}{2})(3x^{2}+4x−1) \)
- \((\frac{1}{3}x−\frac{1}{4})(3x^{2}+9x−3) \)
- \((x+3)^{3}\)
- \((x−2)^{3}\)
- \((3x−1)^{3}\)
- \((2x+y)^{3}\)
- \((5x−2)(2x^{3}−4x^{2}+3x−2)\)
- \((x^{2}−2)(x^{3}−2x^{2}+x+1)\)
- Answer
-
1. \(3x^{2}+10x−8 \)
3. \(x^{2}−1 \)
5. \(2x^{2}+x−15 \)
7. \(−3x^{2}+4x−1 \)
9. \(y^{2}−\frac{4}{9}\)
11. \(\frac{3}{16}x^{2}+\frac{7}{20}x+\frac{2}{25}\)
13. \(y^{3}+2y^{2}−2y−4\)
15. \(a^{4}−b^{4}\)
17. \(2x^{3}−7x^{2}−11x−20 \)
19. \(8x^{3}−27 \)
21. \(3x^{3}+\frac{5}{2}x^{2}−3x+12\)
23. \(x^{3}+9x^{2}+27x+27\)
25. \(27x^{3}−27x^{2}+9x−1 \)
27. \(10x^{4}−24x^{3}+23x^{2}−16x+4\)
Multiply.
- \((x^{2}−x+1)(x^{2}+2x+1)\)
- \((3x^{2}−2x−1)(2x^{2}+3x−4)\)
- \((2x^{2}−3x+5)(x^{2}+5x−1)\)
- \((a+b+c)(a−b−c)\)
- \((a+2b−c)^{2}\)
- \((x+y+z)^{2}\)
- \((x−3)^{4}\)
- \((x+y)^{4}\)
- Find the volume of a rectangular solid with sides measuring \(x, x+2\), and \(x+4\) units.
- Find the volume of a cube where each side measures \(x−5\) units.
- Answer
-
1. \(x^{4}+x^{3}+x+1\)
3. \(2x^{4}+7x^{3}−12x^{2}+28x−5\)
5. \(a^{2}+4ab−2ac+4b^{2}−4bc+c^{2}\)
7. \(x^{4}−12x^{3}+54x^{2}−108x+81\)
9. \(x^{3}+6x^{2}+8x\)
Multiply.
- \((x+2)^{2}\)
- \((x−3)^{2}\)
- \((2x+5)^{2}\)
- \((3x−7)^{2}\)
- \((−x+2)^{2}\)
- \((−9x+1)^{2}\)
- \((a+6)^{2}\)
- \((2a−3b)^{2}\)
- \((\frac{2}{3}x+\frac{3}{4})^{2}\)
- \((\frac{1}{2}x−\frac{3}{5})^{2}\)
- \((x^{2}+2)^{2}\)
- \((x^{2}+y^{2})^{2}\)
- \((x+4)(x−4)\)
- \((2x+1)(2x−1)\)
- \((5x+3)(5x−3)\)
- \((\frac{1}{5}x−\frac{1}{3})(\frac{1}{5}x+\frac{1}{3})\)
- \((\frac{3}{2}x+\frac{2}{5})(\frac{3}{2}x−\frac{2}{5})\)
- \((2x−3y)(2x+3y)\)
- \((4x−y)(4x+y)\)
- \((a^{3}−b^{3})(a^{3}+b^{3})\)
- A box is made by cutting out the corners and folding up the edges of a square piece of cardboard. A template for a cardboard box with a height of \(2\) inches is given. Find a formula for the volume, if the initial piece of cardboard is a square with sides measuring \(x\) inches.
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Figure \(\PageIndex{5}\) - A template for a cardboard box with a height of \(x\) inches is given. Find a formula for the volume, if the initial piece of cardboard is a square with sides measuring \(12\) inches.
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Figure \(\PageIndex{6}\)
- Answer
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1. \(x^{2}+4x+4 \)
3. \(4x^{2}+20x+25 \)
5. \(x^{2}−4x+4 \)
7. \(a^{2}+12a+36 \)
9. \(\frac{4}{9}x^{2}+x+\frac{9}{16}\)
11. \(x^{4}+4x^{2}+4\)
13. \(x^{2}−16 \)
15. \(25x^{2}−9 \)
17. \(\frac{9}{4}x^{2}−\frac{4}{25}\)
19. \(16x^{2}−y^{2}\)
21. \(V=2x^{2}−16x+32\) cubic inches
For each problem, calculate \((f⋅g)(x)\), given the functions.
- \(f(x)=8x\) and \(g(x)=3x−5\)
- \(f(x)=x^{2}\) and \(g(x)=−5x+1\)
- \(f(x)=x−7\) and \(g(x)=6x−1\)
- \(f(x)=5x+3\) and \(g(x)=x^{2}+2x−3\)
- \(f(x)=x^{2}+6x−3\) and \(g(x)=2x^{2}−3x+5\)
- \(f(x)=3x^{2}−x+1\) and \(g(x)=−x^{2}+2x−1\)
- Answer
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1. \((f⋅g)(x)=24x^{2}−40x\)
3. \((f⋅g)(x)=6x^{2}−43x+7\)
5. \((f⋅g)(x)=2x^{4}+9x^{3}−19x^{2}+39x−15\)
Given \(f(x)=2x−3\) and \(g(x)=3x−1\), find the following
- \((f⋅g)(x)\)
- \((g⋅f)(x)\)
- \((f⋅g)(0)\)
- \((f⋅g)(−1)\)
- \((f⋅g)(1)\)
- \((f⋅g)(\frac{1}{2})\)
- Answer
-
1. \((f⋅g)(x)=6x^{2}−11x+3\)
3. \((f⋅g)(0)=3\)
5. \((f⋅g)(1)=−2\)
Given \(f(x)=5x−1\) and \(g(x)=2x^{2}−4x+5\), find the following.
- \((f⋅g)(x) \)
- \((g⋅f)(x) \)
- \((f⋅g)(0) \)
- \((f⋅g)(−1) \)
- \((f⋅g)(1) \)
- \((f⋅g)(\frac{1}{2})\)
- \((f⋅f)(x) \)
- \((g⋅g)(x)\)
- Answer
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1. \((f⋅g)(x)=10x^{3}−22x^{2}+29x−5\)
3. \((f⋅g)(0)=−5\)
5. \((f⋅g)(1)=12\)
7. \((f⋅f)(x)=25x^{2}−10x+1\)
- Explain why \((x+y)^{2}\neq x^{2}+y{2}\).
- Explain how to quickly multiply a binomial with its conjugate. Give an example.
- What are the advantages and disadvantages of using the mnemonic device FOIL?
- Answer
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1. Answers may vary
3. Answers may vary