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

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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}
\begin{gathered} (x\text{ - 3\rparen}^5\text{ = \lparen1\rparen}x^5\text{ + \lparen5\rparen}x^(5-1)y\text{ + \lparen10\rparen}x^(5-2)y^2+\text{ \lparen10\rparen}x^(5-3)y^3\text{ + \lparen5\rparen}x^(5-4)y^4\text{ + \lparen1\rparen}y^5 \\ \\ y\text{ = -3} \\ (x\text{ - 3\rparen}^5\text{ = \lparen1\rparen}x^5\text{ + \lparen5\rparen}* x^4*(-3)\text{ + \lparen10\rparen}* x^3*(-3)^2+\text{ \lparen10\rparen}* x^2*(-3)^3\text{ + \lparen5\rparen}* x^1*(-3)^4\text{ + \lparen1\rparen\lparen-3\rparen}^5 \\ \\ (x\text{ - 3\rparen}^5\text{ = }x^5\text{ -15}x^4\text{ + 90}x^3-27\text{0}x^2\text{ + 405}x^1\text{ -243} \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}

Expand (x-3)^5 using the vi al theorem and Pascal’s triangle. Show all the steps-example-1
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