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If (x + 2)(3x^2 - x + 8) = Ax^3 + Bx^2 + Cx + D, what is the value of A + 2B + 3C + 4D?

 May 4, 2022

Best Answer 

 #1
avatar+9310 
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\(\quad(x + 2)(3x^2 - x + 8)\\ = x(3x^2 - x + 8) + 2(3x^2 - x + 8)\\ = 3x^3 - x^2 + 8x + 6x^2 - 2x + 16\\ = 3x^3 + 5x^2 + 6x + 16\)

 

Now you can compare the coefficients of \(3x^3 + 5x^2 + 6x + 16\) and \(Ax^3 + Bx^2 + Cx + D\) to find A, B, C, D. Then just calculate A + 2B + 3C + 4D.

 May 4, 2022
 #1
avatar+9310 
+1
Best Answer

Expand out. 

 

\(\quad(x + 2)(3x^2 - x + 8)\\ = x(3x^2 - x + 8) + 2(3x^2 - x + 8)\\ = 3x^3 - x^2 + 8x + 6x^2 - 2x + 16\\ = 3x^3 + 5x^2 + 6x + 16\)

 

Now you can compare the coefficients of \(3x^3 + 5x^2 + 6x + 16\) and \(Ax^3 + Bx^2 + Cx + D\) to find A, B, C, D. Then just calculate A + 2B + 3C + 4D.

MaxWong May 4, 2022

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