The Binomial Theorem is a shortcut for expanding powers of binomials such as \((x+y)^n\). It uses binomial coefficients from combinations or Pascal’s Triangle to build each term of the expansion. These problems practice expanding binomials, finding specific terms, finding coefficients, and handling signs carefully when subtraction is involved.
Lesson
Notes
Practice Problems
Expand the following
\(\textbf{1)}\) \((a+b)^4\) The expansion is \(a^4+4a^3 b+6a^2 b^2+4ab^3+b^4\)
\(\,\,\,\,\,\,\text{Use coefficients }1,4,6,4,1\text{ from Pascal’s Triangle.}\)
\(\,\,\,\,\,\,(a+b)^4\)
\(\,\,\,\,\,\,=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
\(\,\,\,\,\,\,\text{Use coefficients }1,4,6,4,1\text{ from Pascal’s Triangle.}\)
\(\,\,\,\,\,\,(a+b)^4\)
\(\,\,\,\,\,\,=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
\(\textbf{2)}\) \((a+2b)^3\) The expansion is \(a^3+6a^2 b+12a b^2+8b^3\)
\(\,\,\,\,\,\,\text{Use coefficients }1,3,3,1.\)
\(\,\,\,\,\,\,(a+2b)^3\)
\(\,\,\,\,\,\,=a^3+3a^2(2b)+3a(2b)^2+(2b)^3\)
\(\,\,\,\,\,\,=a^3+6a^2b+12ab^2+8b^3\)
\(\,\,\,\,\,\,\text{Use coefficients }1,3,3,1.\)
\(\,\,\,\,\,\,(a+2b)^3\)
\(\,\,\,\,\,\,=a^3+3a^2(2b)+3a(2b)^2+(2b)^3\)
\(\,\,\,\,\,\,=a^3+6a^2b+12ab^2+8b^3\)
\(\textbf{3)}\) \((3x+y)^5\) The expansion is \(243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5\)
\(\,\,\,\,\,\,\text{Use coefficients }1,5,10,10,5,1.\)
\(\,\,\,\,\,\,(3x+y)^5\)
\(\,\,\,\,\,\,=(3x)^5+5(3x)^4y+10(3x)^3y^2+10(3x)^2y^3+5(3x)y^4+y^5\)
\(\,\,\,\,\,\,=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5\)
\(\,\,\,\,\,\,\text{Use coefficients }1,5,10,10,5,1.\)
\(\,\,\,\,\,\,(3x+y)^5\)
\(\,\,\,\,\,\,=(3x)^5+5(3x)^4y+10(3x)^3y^2+10(3x)^2y^3+5(3x)y^4+y^5\)
\(\,\,\,\,\,\,=243x^5+405x^4y+270x^3y^2+90x^2y^3+15xy^4+y^5\)
\(\textbf{4)}\) \((3a-b)^4\) The expansion is \(81a^4-108a^3b+54a^2b^2-12ab^3+b^4\)
\(\,\,\,\,\,\,\text{Use coefficients }1,4,6,4,1.\)
\(\,\,\,\,\,\,(3a-b)^4\)
\(\,\,\,\,\,\,=(3a)^4+4(3a)^3(-b)+6(3a)^2(-b)^2+4(3a)(-b)^3+(-b)^4\)
\(\,\,\,\,\,\,=81a^4-108a^3b+54a^2b^2-12ab^3+b^4\)
\(\,\,\,\,\,\,\text{Use coefficients }1,4,6,4,1.\)
\(\,\,\,\,\,\,(3a-b)^4\)
\(\,\,\,\,\,\,=(3a)^4+4(3a)^3(-b)+6(3a)^2(-b)^2+4(3a)(-b)^3+(-b)^4\)
\(\,\,\,\,\,\,=81a^4-108a^3b+54a^2b^2-12ab^3+b^4\)
\(\textbf{5)}\) \((x-2y)^5\) The expansion is \(x^5-10x^4 y+40x^3 y^2-80x^2 y^3+80xy^4-32y^5\)
\(\,\,\,\,\,\,\text{Use coefficients }1,5,10,10,5,1.\)
\(\,\,\,\,\,\,(x-2y)^5\)
\(\,\,\,\,\,\,=x^5+5x^4(-2y)+10x^3(-2y)^2+10x^2(-2y)^3+5x(-2y)^4+(-2y)^5\)
\(\,\,\,\,\,\,=x^5-10x^4y+40x^3y^2-80x^2y^3+80xy^4-32y^5\)
\(\,\,\,\,\,\,\text{Use coefficients }1,5,10,10,5,1.\)
\(\,\,\,\,\,\,(x-2y)^5\)
\(\,\,\,\,\,\,=x^5+5x^4(-2y)+10x^3(-2y)^2+10x^2(-2y)^3+5x(-2y)^4+(-2y)^5\)
\(\,\,\,\,\,\,=x^5-10x^4y+40x^3y^2-80x^2y^3+80xy^4-32y^5\)
\(\textbf{6)}\) \((x-y)^6\) The expansion is \(x^6-6x^5y+15x^4y^2-20x^3y^3+15x^2y^4-6xy^5+y^6\)
\(\,\,\,\,\,\,\text{Use coefficients }1,6,15,20,15,6,1.\)
\(\,\,\,\,\,\,(x-y)^6\)
\(\,\,\,\,\,\,=x^6+6x^5(-y)+15x^4(-y)^2+20x^3(-y)^3+15x^2(-y)^4+6x(-y)^5+(-y)^6\)
\(\,\,\,\,\,\,=x^6-6x^5y+15x^4y^2-20x^3y^3+15x^2y^4-6xy^5+y^6\)
\(\,\,\,\,\,\,\text{Use coefficients }1,6,15,20,15,6,1.\)
\(\,\,\,\,\,\,(x-y)^6\)
\(\,\,\,\,\,\,=x^6+6x^5(-y)+15x^4(-y)^2+20x^3(-y)^3+15x^2(-y)^4+6x(-y)^5+(-y)^6\)
\(\,\,\,\,\,\,=x^6-6x^5y+15x^4y^2-20x^3y^3+15x^2y^4-6xy^5+y^6\)
\(\textbf{7)}\) Find the \(6th\) term of \((x-2)^{10}\) The \(6th\) term is \(-8064x^5\)
\(\,\,\,\,\,\,T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
\(\,\,\,\,\,\,\text{For the 6th term, }r=5.\)
\(\,\,\,\,\,\,T_6=\binom{10}{5}x^{10-5}(-2)^5\)
\(\,\,\,\,\,\,T_6=252x^5(-32)\)
\(\,\,\,\,\,\,T_6=-8064x^5\)
\(\,\,\,\,\,\,T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
\(\,\,\,\,\,\,\text{For the 6th term, }r=5.\)
\(\,\,\,\,\,\,T_6=\binom{10}{5}x^{10-5}(-2)^5\)
\(\,\,\,\,\,\,T_6=252x^5(-32)\)
\(\,\,\,\,\,\,T_6=-8064x^5\)
\(\textbf{8)}\) Find the \(5th\) term of \((2x+y)^8\) The \(5th\) term is \(1120x^4 y^4\)
\(\,\,\,\,\,\,T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
\(\,\,\,\,\,\,\text{For the 5th term, }r=4.\)
\(\,\,\,\,\,\,T_5=\binom{8}{4}(2x)^{8-4}y^4\)
\(\,\,\,\,\,\,T_5=70(2x)^4y^4\)
\(\,\,\,\,\,\,T_5=70(16x^4)y^4\)
\(\,\,\,\,\,\,T_5=1120x^4y^4\)
\(\,\,\,\,\,\,T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
\(\,\,\,\,\,\,\text{For the 5th term, }r=4.\)
\(\,\,\,\,\,\,T_5=\binom{8}{4}(2x)^{8-4}y^4\)
\(\,\,\,\,\,\,T_5=70(2x)^4y^4\)
\(\,\,\,\,\,\,T_5=70(16x^4)y^4\)
\(\,\,\,\,\,\,T_5=1120x^4y^4\)
\(\textbf{9)}\) Find the \(3rd\) term of \((x+3y)^6\) The \(3rd\) term is \(135x^4 y^2\)
\(\,\,\,\,\,\,T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
\(\,\,\,\,\,\,\text{For the 3rd term, }r=2.\)
\(\,\,\,\,\,\,T_3=\binom{6}{2}x^{6-2}(3y)^2\)
\(\,\,\,\,\,\,T_3=15x^4(9y^2)\)
\(\,\,\,\,\,\,T_3=135x^4y^2\)
\(\,\,\,\,\,\,T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
\(\,\,\,\,\,\,\text{For the 3rd term, }r=2.\)
\(\,\,\,\,\,\,T_3=\binom{6}{2}x^{6-2}(3y)^2\)
\(\,\,\,\,\,\,T_3=15x^4(9y^2)\)
\(\,\,\,\,\,\,T_3=135x^4y^2\)
\(\textbf{10)}\) Find the \(5th\) term of \((x-2y)^9\) The \(5th\) term is \(2016x^5 y^4\)
\(\,\,\,\,\,\,T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
\(\,\,\,\,\,\,\text{For the 5th term, }r=4.\)
\(\,\,\,\,\,\,T_5=\binom{9}{4}x^{9-4}(-2y)^4\)
\(\,\,\,\,\,\,T_5=126x^5(16y^4)\)
\(\,\,\,\,\,\,T_5=2016x^5y^4\)
\(\,\,\,\,\,\,T_{r+1}=\binom{n}{r}a^{n-r}b^r\)
\(\,\,\,\,\,\,\text{For the 5th term, }r=4.\)
\(\,\,\,\,\,\,T_5=\binom{9}{4}x^{9-4}(-2y)^4\)
\(\,\,\,\,\,\,T_5=126x^5(16y^4)\)
\(\,\,\,\,\,\,T_5=2016x^5y^4\)
\(\textbf{11)}\) \((x+3)^4\) The expansion is \(x^4+12x^3+54x^2+108x+81\)
\(\,\,\,\,\,\,\text{Use coefficients }1,4,6,4,1.\)
\(\,\,\,\,\,\,(x+3)^4\)
\(\,\,\,\,\,\,=x^4+4x^3(3)+6x^2(3)^2+4x(3)^3+3^4\)
\(\,\,\,\,\,\,=x^4+12x^3+54x^2+108x+81\)
\(\,\,\,\,\,\,\text{Use coefficients }1,4,6,4,1.\)
\(\,\,\,\,\,\,(x+3)^4\)
\(\,\,\,\,\,\,=x^4+4x^3(3)+6x^2(3)^2+4x(3)^3+3^4\)
\(\,\,\,\,\,\,=x^4+12x^3+54x^2+108x+81\)
\(\textbf{12)}\) \((2x-1)^5\) The expansion is \(32x^5-80x^4+80x^3-40x^2+10x-1\)
\(\,\,\,\,\,\,\text{Use coefficients }1,5,10,10,5,1.\)
\(\,\,\,\,\,\,(2x-1)^5\)
\(\,\,\,\,\,\,=(2x)^5+5(2x)^4(-1)+10(2x)^3(-1)^2+10(2x)^2(-1)^3+5(2x)(-1)^4+(-1)^5\)
\(\,\,\,\,\,\,=32x^5-80x^4+80x^3-40x^2+10x-1\)
\(\,\,\,\,\,\,\text{Use coefficients }1,5,10,10,5,1.\)
\(\,\,\,\,\,\,(2x-1)^5\)
\(\,\,\,\,\,\,=(2x)^5+5(2x)^4(-1)+10(2x)^3(-1)^2+10(2x)^2(-1)^3+5(2x)(-1)^4+(-1)^5\)
\(\,\,\,\,\,\,=32x^5-80x^4+80x^3-40x^2+10x-1\)
\(\textbf{13)}\) \((m-2n)^4\) The expansion is \(m^4-8m^3n+24m^2n^2-32mn^3+16n^4\)
\(\,\,\,\,\,\,\text{Use coefficients }1,4,6,4,1.\)
\(\,\,\,\,\,\,(m-2n)^4\)
\(\,\,\,\,\,\,=m^4+4m^3(-2n)+6m^2(-2n)^2+4m(-2n)^3+(-2n)^4\)
\(\,\,\,\,\,\,=m^4-8m^3n+24m^2n^2-32mn^3+16n^4\)
\(\,\,\,\,\,\,\text{Use coefficients }1,4,6,4,1.\)
\(\,\,\,\,\,\,(m-2n)^4\)
\(\,\,\,\,\,\,=m^4+4m^3(-2n)+6m^2(-2n)^2+4m(-2n)^3+(-2n)^4\)
\(\,\,\,\,\,\,=m^4-8m^3n+24m^2n^2-32mn^3+16n^4\)
\(\textbf{14)}\) Find the coefficient of \(x^3y^4\) in \((2x-y)^7\) The coefficient is \(280\)
\(\,\,\,\,\,\,T_{r+1}=\binom{7}{r}(2x)^{7-r}(-y)^r\)
\(\,\,\,\,\,\,\text{To get }y^4,\text{ use }r=4.\)
\(\,\,\,\,\,\,T_5=\binom{7}{4}(2x)^3(-y)^4\)
\(\,\,\,\,\,\,T_5=35(8x^3)y^4\)
\(\,\,\,\,\,\,T_5=280x^3y^4\)
\(\,\,\,\,\,\,\text{The coefficient is }280.\)
\(\,\,\,\,\,\,T_{r+1}=\binom{7}{r}(2x)^{7-r}(-y)^r\)
\(\,\,\,\,\,\,\text{To get }y^4,\text{ use }r=4.\)
\(\,\,\,\,\,\,T_5=\binom{7}{4}(2x)^3(-y)^4\)
\(\,\,\,\,\,\,T_5=35(8x^3)y^4\)
\(\,\,\,\,\,\,T_5=280x^3y^4\)
\(\,\,\,\,\,\,\text{The coefficient is }280.\)
\(\textbf{15)}\) Find the coefficient of \(x^6\) in \((x+2)^8\) The coefficient is \(112\)
\(\,\,\,\,\,\,T_{r+1}=\binom{8}{r}x^{8-r}(2)^r\)
\(\,\,\,\,\,\,\text{To get }x^6,\text{ solve }8-r=6.\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,T_3=\binom{8}{2}x^6(2)^2\)
\(\,\,\,\,\,\,T_3=28x^6(4)=112x^6\)
\(\,\,\,\,\,\,\text{The coefficient is }112.\)
\(\,\,\,\,\,\,T_{r+1}=\binom{8}{r}x^{8-r}(2)^r\)
\(\,\,\,\,\,\,\text{To get }x^6,\text{ solve }8-r=6.\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,T_3=\binom{8}{2}x^6(2)^2\)
\(\,\,\,\,\,\,T_3=28x^6(4)=112x^6\)
\(\,\,\,\,\,\,\text{The coefficient is }112.\)
Challenge Problems
\(\textbf{16)}\) Find the coefficient of \(x^4\) in \((3x-2)^6\) The coefficient is \(1215\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}(3x)^{6-r}(-2)^r\)
\(\,\,\,\,\,\,\text{To get }x^4,\text{ solve }6-r=4.\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,T_3=\binom{6}{2}(3x)^4(-2)^2\)
\(\,\,\,\,\,\,T_3=15(81x^4)(4)\)
\(\,\,\,\,\,\,T_3=4860x^4\)
\(\,\,\,\,\,\,\text{The coefficient is }4860.\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}(3x)^{6-r}(-2)^r\)
\(\,\,\,\,\,\,\text{To get }x^4,\text{ solve }6-r=4.\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,T_3=\binom{6}{2}(3x)^4(-2)^2\)
\(\,\,\,\,\,\,T_3=15(81x^4)(4)\)
\(\,\,\,\,\,\,T_3=4860x^4\)
\(\,\,\,\,\,\,\text{The coefficient is }4860.\)
\(\textbf{17)}\) Find the middle term of \((x+2y)^8\) The middle term is \(1120x^4y^4\)
\(\,\,\,\,\,\,\text{There are }8+1=9\text{ terms, so the middle term is the 5th term.}\)
\(\,\,\,\,\,\,\text{For the 5th term, }r=4.\)
\(\,\,\,\,\,\,T_5=\binom{8}{4}x^{8-4}(2y)^4\)
\(\,\,\,\,\,\,T_5=70x^4(16y^4)\)
\(\,\,\,\,\,\,T_5=1120x^4y^4\)
\(\,\,\,\,\,\,\text{There are }8+1=9\text{ terms, so the middle term is the 5th term.}\)
\(\,\,\,\,\,\,\text{For the 5th term, }r=4.\)
\(\,\,\,\,\,\,T_5=\binom{8}{4}x^{8-4}(2y)^4\)
\(\,\,\,\,\,\,T_5=70x^4(16y^4)\)
\(\,\,\,\,\,\,T_5=1120x^4y^4\)
\(\textbf{18)}\) Find the constant term in \(\left(x+\frac{2}{x}\right)^6\) The constant term is \(160\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}x^{6-r}\left(\frac{2}{x}\right)^r\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}2^r x^{6-r-r}\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}2^r x^{6-2r}\)
\(\,\,\,\,\,\,\text{For the constant term, }6-2r=0.\)
\(\,\,\,\,\,\,r=3\)
\(\,\,\,\,\,\,T_4=\binom{6}{3}2^3\)
\(\,\,\,\,\,\,T_4=20(8)=160\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}x^{6-r}\left(\frac{2}{x}\right)^r\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}2^r x^{6-r-r}\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}2^r x^{6-2r}\)
\(\,\,\,\,\,\,\text{For the constant term, }6-2r=0.\)
\(\,\,\,\,\,\,r=3\)
\(\,\,\,\,\,\,T_4=\binom{6}{3}2^3\)
\(\,\,\,\,\,\,T_4=20(8)=160\)
\(\textbf{19)}\) Find the term containing \(x^2\) in \(\left(2x-\frac{1}{x}\right)^6\) The term is \(240x^2\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}(2x)^{6-r}\left(-\frac{1}{x}\right)^r\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}2^{6-r}(-1)^r x^{6-r-r}\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}2^{6-r}(-1)^r x^{6-2r}\)
\(\,\,\,\,\,\,\text{Set }6-2r=2.\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,T_3=\binom{6}{2}2^4(-1)^2x^2\)
\(\,\,\,\,\,\,T_3=15(16)x^2\)
\(\,\,\,\,\,\,T_3=240x^2\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}(2x)^{6-r}\left(-\frac{1}{x}\right)^r\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}2^{6-r}(-1)^r x^{6-r-r}\)
\(\,\,\,\,\,\,T_{r+1}=\binom{6}{r}2^{6-r}(-1)^r x^{6-2r}\)
\(\,\,\,\,\,\,\text{Set }6-2r=2.\)
\(\,\,\,\,\,\,r=2\)
\(\,\,\,\,\,\,T_3=\binom{6}{2}2^4(-1)^2x^2\)
\(\,\,\,\,\,\,T_3=15(16)x^2\)
\(\,\,\,\,\,\,T_3=240x^2\)
\(\textbf{20)}\) Find the coefficient of \(x^5y^3\) in \((2x-3y)^8\) The coefficient is \(-12096\)
\(\,\,\,\,\,\,T_{r+1}=\binom{8}{r}(2x)^{8-r}(-3y)^r\)
\(\,\,\,\,\,\,\text{To get }y^3,\text{ use }r=3.\)
\(\,\,\,\,\,\,T_4=\binom{8}{3}(2x)^5(-3y)^3\)
\(\,\,\,\,\,\,T_4=56(32x^5)(-27y^3)\)
\(\,\,\,\,\,\,T_4=-48384x^5y^3\)
\(\,\,\,\,\,\,\text{The coefficient is }-48384.\)
\(\,\,\,\,\,\,T_{r+1}=\binom{8}{r}(2x)^{8-r}(-3y)^r\)
\(\,\,\,\,\,\,\text{To get }y^3,\text{ use }r=3.\)
\(\,\,\,\,\,\,T_4=\binom{8}{3}(2x)^5(-3y)^3\)
\(\,\,\,\,\,\,T_4=56(32x^5)(-27y^3)\)
\(\,\,\,\,\,\,T_4=-48384x^5y^3\)
\(\,\,\,\,\,\,\text{The coefficient is }-48384.\)
See Related Pages\(\)
\(\bullet\text{ Multiplying Polynomials}\)
\(\,\,\,\,\,\,\,\,(x+2)(x^2+3x−5)…\)
\(\bullet\text{ Geometric Sequences}\)
\(\,\,\,\,\,\,\,a_n=a_1 \cdot r^{(n-1)}…\)
\(\bullet\text{ Fibonacci Sequence}\)
\(\,\,\,\,\,\,\,\,0,1,1,2,3,5,8,13,21…\)
\(\bullet\text{ Golden Ratio}\)
\(\,\,\,\,\,\,\,\,1.61803398875…\)
\(\bullet\text{ Pascal’s Triangle}\)
\(\,\,\,\,\,\,\,\,\)