Solveeit Logo
Login

Question

Question: How do you use Pascal’s Triangle to expand \({\left( {x - 2} \right)^4}\)?...

How do you use Pascal’s Triangle to expand (x−2)4{\left( {x - 2} \right)^4}?

Explanation

Solution

We will first form the Pascal’s Triangle up to the power of 4, then see the numbers as the coefficient of xmyn{x^m}{y^n} where we keep on decreasing m and increasing n. Now, we will just put in the value of y as – 2 to find the answer.

Complete step by step solution:
We are given that we are required to use the Pascal’s triangle to expand (x−2)4{\left( {x - 2} \right)^4}.
We know that the Pascal’ Triangle is given by the following figure:-

| | | | | 1| | | |
---|---|---|---|---|---|---|---|---|---
| | | | 1| | 1| | |
x2{x^2}| | | 1| | 2| | 1| |
x3{x^3}| | 1| | 3| | 3| | 1|
x4{x^4}| 1| | 4| | 6| | 4| | 1

This is Pascal's triangle.
Now, we see that in x4{x^4}, we have 1, 4, 6, 4 and 1.
This means that (x+y)4=1×(x4y0)+4×(x3y1)+6×(x2y2)+4×(x1y3)+1×(x0y4){\left( {x + y} \right)^4} = 1 \times \left( {{x^4}{y^0}} \right) + 4 \times \left( {{x^3}{y^1}} \right) + 6 \times \left( {{x^2}{y^2}} \right) + 4 \times \left( {{x^1}{y^3}} \right) + 1 \times \left( {{x^0}{y^4}} \right)
Now, we will just simplify it to obtain the following expression with us:-
⇒(x+y)4=x4+4x3y+6x2y2+4xy3+y4\Rightarrow {\left( {x + y} \right)^4} = {x^4} + 4{x^3}y + 6{x^2}{y^2} + 4x{y^3} + {y^4}
Now, let us put in y=−2y = - 2 in the above mentioned expression so that we obtain the following expression with us:-
⇒(x−2)4=x4+4x3(−2)+6x2(−2)2+4x(−2)3+(−2)4\Rightarrow {\left( {x - 2} \right)^4} = {x^4} + 4{x^3}( - 2) + 6{x^2}{( - 2)^2} + 4x{( - 2)^3} + {( - 2)^4}
Simplifying the terms on the right hand side of the above expression, we will then obtain the following expression with us:-
⇒(x−2)4=x4+4(−2)x3+6(4)x2+4(−8)x+16\Rightarrow {\left( {x - 2} \right)^4} = {x^4} + 4( - 2){x^3} + 6(4){x^2} + 4( - 8)x + 16
Simplifying the terms on the right hand side further of the above expression, we will then obtain the following expression with us:-
⇒(x−2)4=x4−8x3+24x2−32x+16\Rightarrow {\left( {x - 2} \right)^4} = {x^4} - 8{x^3} + 24{x^2} - 32x + 16
Thus we have the required answer.

Note:
The students must note that if not mentioned that we need to expand it by Pascal’s triangle, we may use some other ways to solve it:
Alternate Way:
We are given that we are required to use the Pascal’s triangle to expand (x−2)4{\left( {x - 2} \right)^4}.
We can write it as ((x−2)2)2{\left( {{{\left( {x - 2} \right)}^2}} \right)^2}.
Now, we know that we have a formula given by the following expression:-
⇒(a−b)2=a2+b2−2ab\Rightarrow {(a - b)^2} = {a^2} + {b^2} - 2ab
Replacing a by x and b by – 2, we will then obtain the following expression:-
⇒(x−2)2=x2+4−4x\Rightarrow {(x - 2)^2} = {x^2} + 4 - 4x
Now, we know that we have a formula given by the following expression:-
⇒(a+b+c)2=a2+b2+c2+2ab+2bc+2ca\Rightarrow {(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca
Putting a as x2{x^2}, b as 4 and c as −4x - 4x, we will then obtain the following expression:-
⇒(x2+4−4x)2=x4+16+16x2+8x2−32x−8x3\Rightarrow {\left( {{x^2} + 4 - 4x} \right)^2} = {x^4} + 16 + 16{x^2} + 8{x^2} - 32x - 8{x^3}
Simplifying it by combining the like terms, we will then obtain the following expression:-
⇒(x2+4−4x)2=x4−8x3+24x2−32x+16\Rightarrow {\left( {{x^2} + 4 - 4x} \right)^2} = {x^4} - 8{x^3} + 24{x^2} - 32x + 16