Question

Expand (x-3)^5 using the vi al theorem and Pascal’s triangle. Show all the steps

Answer 1

Answer:
`(x\text{ - 3\rparen}^5\text{ = }x^5\text{ - 15}x^4\text{ + 90}x^3-27\text{0}x^2\text{ + 405}x-\text{243}`

Step-by-step explanation:

Given:

`(x\text{ - 3\rparen}^5`

To find:

To expand the expression using binomial's theorem

To expand, we will apply the binomial theorem formula:

`(x\text{ + y\rparen}^n=\text{ }^nC_0x^n\text{ + }^nC_1x^(n-1)y\text{ + }^nC_2x^(n-2)y^2+.\text{ . . + }^nC_ny^n`
`\begin{gathered} (x\text{ - 3\rparen}^5\text{ = }^5C_0x^5\text{ + }^5C_1x^(5-1)y\text{ + }^5C_2x^(5-2)y^2+\text{ }^5C_3x^(5-3)y^3\text{ + }^5C_4x^(5-4)y^4\text{ + }^5C_5y^5 \\ \\ We\text{ will use pascal's triangle to get the coefficient of the terms.} \\ The\text{ coefficients represents the value of the combinations} \\ \\ The\text{ pascals triangle coefficients for power of 5 = 1 5 10 10 5 1} \end{gathered}`

The expansion becomes:

`(x\text{ - 3\rparen}^5\text{ = }x^5\text{ - 15}x^4\text{ + 90}x^3-27\text{0}x^2\text{ + 405x - 243}`