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Multiplying Polynomials by Monomial

Related calculator: Polynomial Calculator

To multiply polynomial by monomial, one should use distributive property of multiplication.

Then, just multiply monomials and you're done.

Example 1. Multiply (2x)(3x2+5x+4){\left({2}{x}\right)}{\left({3}{{x}}^{{2}}+{5}{x}+{4}\right)}.

(2x)(3x2+5x+4)={\color{red}{{{\left({2}{x}\right)}}}}{\color{green}{{{\left({3}{{x}}^{{2}}+{5}{x}+{4}\right)}}}}=

=(2x)⋅(3x2)+(2x)⋅(5x)+(2x)⋅(4)=={\color{red}{{{\left({2}{x}\right)}}}}\cdot{\color{green}{{{\left({3}{{x}}^{{2}}\right)}}}}+{\color{red}{{{\left({2}{x}\right)}}}}\cdot{\color{green}{{{\left({5}{x}\right)}}}}+{\color{red}{{{\left({2}{x}\right)}}}}\cdot{\color{green}{{{\left({4}\right)}}}}= (distributive property of multiplication)

=6x3+10x2+8x={6}{{x}}^{{3}}+{10}{{x}}^{{2}}+{8}{x} (multiply monomials)

Answer: (2x)(3x2+5x+4)=6x3+10x2+8x{\left({2}{x}\right)}{\left({3}{{x}}^{{2}}+{5}{x}+{4}\right)}={6}{{x}}^{{3}}+{10}{{x}}^{{2}}+{8}{x}.

Negative terms are handled in the same way.

Example 2. Multiply the following: (x3−5x2−x+7)13x2{\left({{x}}^{{3}}-{5}{{x}}^{{2}}-{x}+{7}\right)}\frac{{1}}{{3}}{{x}}^{{2}}.

(x3−5x2−x+7)13x2={\color{green}{{{\left({{x}}^{{3}}-{5}{{x}}^{{2}}-{x}+{7}\right)}}}}{\color{red}{{\frac{{1}}{{3}}{{x}}^{{2}}}}}=

=x3⋅13x2+(−5x2)⋅13x2+(−x)⋅13x2+7⋅13x2=={\color{green}{{{{x}}^{{3}}}}}\cdot{\color{red}{{\frac{{1}}{{3}}{{x}}^{{2}}}}}+{\color{green}{{{\left(-{5}{{x}}^{{2}}\right)}}}}\cdot{\color{red}{{\frac{{1}}{{3}}{{x}}^{{2}}}}}+{\color{green}{{{\left(-{x}\right)}}}}\cdot{\color{red}{{\frac{{1}}{{3}}{{x}}^{{2}}}}}+{\color{green}{{{7}}}}\cdot{\color{red}{{\frac{{1}}{{3}}{{x}}^{{2}}}}}= (distributive property of multiplication)

=13x5−53x4−13x3+73x2=\frac{{1}}{{3}}{{x}}^{{5}}-\frac{{5}}{{3}}{{x}}^{{4}}-\frac{{1}}{{3}}{{x}}^{{3}}+\frac{{7}}{{3}}{{x}}^{{2}} (multiply monomials)

Answer: (x3−5x2−x)13x2=13x5−53x4−13x3+73x2{\left({{x}}^{{3}}-{5}{{x}}^{{2}}-{x}\right)}\frac{{1}}{{3}}{{x}}^{{2}}=\frac{{1}}{{3}}{{x}}^{{5}}-\frac{{5}}{{3}}{{x}}^{{4}}-\frac{{1}}{{3}}{{x}}^{{3}}+\frac{{7}}{{3}}{{x}}^{{2}}.

Of course, polynomials with many variables can also be handled in a similar way.

Example 3. Multiply −3xy2-{3}{x}{{y}}^{{2}} by (3x2y+2xz−5xy2−z){\left({3}{{x}}^{{2}}{y}+{2}{x}{z}-{5}{x}{{y}}^{{2}}-{z}\right)}.

−3xy2(3x2y+2xz−5xy2−z)=-{3}{x}{{y}}^{{2}}{\left({3}{{x}}^{{2}}{y}+{2}{x}{z}-{5}{x}{{y}}^{{2}}-{z}\right)}=

=(−3xy2)(3x2y)+(−3xy2)(2xz)+(−3xy2)(−5xy2)+(−3xy2)(−z)=={\left(-{3}{x}{{y}}^{{2}}\right)}{\left({3}{{x}}^{{2}}{y}\right)}+{\left(-{3}{x}{{y}}^{{2}}\right)}{\left({2}{x}{z}\right)}+{\left(-{3}{x}{{y}}^{{2}}\right)}{\left(-{5}{x}{{y}}^{{2}}\right)}+{\left(-{3}{x}{{y}}^{{2}}\right)}{\left(-{z}\right)}=

=−9x3y3−6x2y2z+15x2y4+3xy2z=-{9}{{x}}^{{3}}{{y}}^{{3}}-{6}{{x}}^{{2}}{{y}}^{{2}}{z}+{15}{{x}}^{{2}}{{y}}^{{4}}+{3}{x}{{y}}^{{2}}{z}.

Answer: −3xy2(3x2y+2xz−5xy2−z)=−9x3y3−6x2y2z+15x2y4+3xy2z-{3}{x}{{y}}^{{2}}{\left({3}{{x}}^{{2}}{y}+{2}{x}{z}-{5}{x}{{y}}^{{2}}-{z}\right)}=-{9}{{x}}^{{3}}{{y}}^{{3}}-{6}{{x}}^{{2}}{{y}}^{{2}}{z}+{15}{{x}}^{{2}}{{y}}^{{4}}+{3}{x}{{y}}^{{2}}{z}.

Now, it is time to exercise.

Exercise 1. Multiply (x3+2x+4)⋅(5x2){\left({{x}}^{{3}}+{2}{x}+{4}\right)}\cdot{\left({5}{{x}}^{{2}}\right)}.

Answer: 5x5+10x3+20x2{5}{{x}}^{{5}}+{10}{{x}}^{{3}}+{20}{{x}}^{{2}}.

Exercise 2. Multiply −27a3(a3−2a2+7b)-\frac{{2}}{{7}}{{a}}^{{3}}{\left({{a}}^{{3}}-{2}{{a}}^{{2}}+{7}{b}\right)}.

Answer: −27a6+47a5−2a3b-\frac{{2}}{{7}}{{a}}^{{6}}+\frac{{4}}{{7}}{{a}}^{{5}}-{2}{{a}}^{{3}}{b}.

Exercise 3. Multiply (−3ab)(5a2b−3a3bc+35a2b2−110ab){\left(-{3}{a}{b}\right)}{\left({5}{{a}}^{{2}}{b}-{3}{{a}}^{{3}}{b}{c}+\frac{{3}}{{5}}{{a}}^{{2}}{{b}}^{{2}}-\frac{{1}}{{10}}{a}{b}\right)}.

Answer: −15a3b2+9a4b2c−95a3b3+310a2b2-{15}{{a}}^{{3}}{{b}}^{{2}}+{9}{{a}}^{{4}}{{b}}^{{2}}{c}-\frac{{9}}{{5}}{{a}}^{{3}}{{b}}^{{3}}+\frac{{3}}{{10}}{{a}}^{{2}}{{b}}^{{2}}.