Question

If (3x-1)^4=a4x^4+a3x^3+a2x^2+a1x0+a0,find the value of a4+3a3+9a2+27a1+81a0

Answer 1

The value of $a_4+3a_3+9a_2+27a_1+81a_0$ is 0.

Step-by-step explanation:

We are given with the following situation below;

If $(3x-1)^{4}=a_4x^{4}+a_3x^{3} +a_2x^{2} +a_1x+a_0$ , then we have to find the value of $a_4+3a_3+9a_2+27a_1+81a_0$.

Firstly, taking the left-hand side of the equation;

$(3x-1)^{4}$ =  $(3x-1)^{2} \times (3x-1)^{2}$

Now, as we know that $(a-b)^{2} = a^{2} +b^{2} -2ab$, so using this we get;

              =  $((3x)^{2} +1^{2} - 2(3x)(1)) \times ((3x)^{2} +1^{2} - 2(3x)(1))$

              =  $(9x^{2} +1 - 6x) \times (9x^{2} +1 - 6x)$

              =  $81x^{4}+9x^{2} - 54x^{3} +9x^{2} +1-6x-54x^{3} -6x+36x^{2}$

              =  $81x^{4} - 108x^{3} +54x^{2} -12x+1$

Now, comparing the result with the expression $a_4x^{4}+a_3x^{3} +a_2x^{2} +a_1x+a_0$, we get;

$a_4$ = 81, $a_3$ = -108, $a_2$ = 54, $a_1$ = 12, and $a_0$ = 1.

So, the value of $a_4+3a_3+9a_2+27a_1+81a_0$ is given by;

      =  $81+3\times (-108)+9\times 54+27\times (-12)+81\times 1$

      =  81 - 324 + 486 - 324 + 81

      =  648 - 648 = 0

Hence, the value of $a_4+3a_3+9a_2+27a_1+81a_0$ is 0.

Answer 2

Answer:

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Step-by-step explanation:

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