Q. Use Pascal's Triangle to expand (3z−2y)5. Express your answer in simplest form.Answer:
Identify Row: Identify the row of Pascal's Triangle that corresponds to the exponent 5. The 6th row (since we start counting from the 0th row for the exponent 0) of Pascal's Triangle is 1,5,10,10,5,1. These numbers will be the coefficients in the expanded form.
Write Expansion Terms: Write out the terms of the expansion using the binomial theorem and the coefficients from Pascal's Triangle.The binomial theorem states that (a+b)n=Σ((kn))∗a(n−k)∗bk, where Σ denotes the sum over k from 0 to n.For (3z−2y)5, the expanded form will be:1∗(3z)5∗(−2y)0+5∗(3z)4∗(−2y)1+10∗(3z)3∗(−2y)2+10∗(3z)2∗(−2y)3+5∗(3z)1∗(−2y)4+1∗(3z)0∗(−2y)5
Simplify Each Term: Simplify each term of the expansion.Now we will simplify each term by calculating the powers and multiplying the coefficients:1×(243z5)(1)+5×(81z4)(−2y)+10×(27z3)(4y2)+10×(9z2)(−8y3)+5×(3z)(16y4)+1×(1)(−32y5)
Combine Coefficients: Combine the coefficients and simplify the terms.Now we multiply the coefficients together for each term:(243z5)+(−810z4y)+(1080z3y2)+(−720z2y3)+(240zy4)+(−32y5)
Write Final Form: Write the final expanded form in simplest terms.The final expanded form of (3z−2y)5 is:243z5−810z4y+1080z3y2−720z2y3+240zy4−32y5
More problems from Pascal's triangle and the Binomial Theorem