Question

Expand \( (2x-3y)^5 \) using the Binomial Theorem.

Hint:

For \((a-b)^n\), the signs alternate: \[ +,-,+,-,+,\dots \] Always use the general term \[ T_{r+1}=\binom{n}{r}a^{n-r}b^r \] to systematically compute each term.

Answer 1

Concept: The Binomial Theorem states that \[ (a+b)^n=\sum_{r=0}^{n} \binom{n}{r} a^{\,n-r} b^{\,r} \] For the expression \( (2x-3y)^5 \), we take \[ a=2x,\qquad b=-3y,\qquad n=5 \]
Step 1: Write the general term. \[ T_{r+1}=\binom{5}{r}(2x)^{5-r}(-3y)^r \] where \(r=0,1,2,3,4,5\).
Step 2: Compute each term. \[ T_1=\binom{5}{0}(2x)^5=32x^5 \] \[ T_2=\binom{5}{1}(2x)^4(-3y) =5\cdot16x^4(-3y) =-240x^4y \] \[ T_3=\binom{5}{2}(2x)^3(9y^2) =10\cdot8x^3\cdot9y^2 =720x^3y^2 \] \[ T_4=\binom{5}{3}(2x)^2(-27y^3) =10\cdot4x^2(-27y^3) =-1080x^2y^3 \] \[ T_5=\binom{5}{4}(2x)(81y^4) =5\cdot2x\cdot81y^4 =810xy^4 \] \[ T_6=\binom{5}{5}(-243y^5) =-243y^5 \]
Step 3: Write the complete expansion. \[ (2x-3y)^5 = 32x^5 -240x^4y +720x^3y^2 -1080x^2y^3 +810xy^4 -243y^5 \]
Step 4: Final Answer. \[ \boxed{ (2x-3y)^5 = 32x^5 -240x^4y +720x^3y^2 -1080x^2y^3 +810xy^4 -243y^5 } \]