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If (x + 2)(3x^2 + x + 5) = Ax^3 + Bx^2 + Cx + D$, what is the value of A + B + 2C + 5D?

 Jun 14, 2022

Best Answer 

 #1
avatar+556 
+1

\(\left(x+2\right)\left(3x^2+x+5\right)\)

 

\(\mathrm{Distribute\:parentheses}\)

 

\(x\cdot \:3x^2+xx+x\cdot \:5+2\cdot \:3x^2+2x+2\cdot \:5\)

 

\(3x^3+7x^2+7x+10\)

 

Sub these in \(A + B + 2C + 5D\)

 

\(3+3+2(7)+5(10)\)

 

\(=70\)

 

-Vinculum

 

smileysmileysmiley

 Jun 14, 2022
 #1
avatar+556 
+1
Best Answer

\(\left(x+2\right)\left(3x^2+x+5\right)\)

 

\(\mathrm{Distribute\:parentheses}\)

 

\(x\cdot \:3x^2+xx+x\cdot \:5+2\cdot \:3x^2+2x+2\cdot \:5\)

 

\(3x^3+7x^2+7x+10\)

 

Sub these in \(A + B + 2C + 5D\)

 

\(3+3+2(7)+5(10)\)

 

\(=70\)

 

-Vinculum

 

smileysmileysmiley

Vinculum Jun 14, 2022
 #3
avatar
+1

Think again, Vinculum

Guest Jun 14, 2022
 #4
avatar+556 
+1

Edit : 3+7+14+50

 

its 74

Vinculum  Jun 14, 2022
 #2
avatar
+1

$$\displaystyle 3x^3 + x^2 + 5x + 6x^2 + 2x + 10, which \ equals \ 3x^3 + 7x^2 + 7x + 10. The \ answer \ is \ 3 + 1 + 14 + 50, which \ is \ 68.$$

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 Jun 14, 2022
 #5
avatar+556 
+1

You mean 3+7+14+50? not 3+1+14+50...

 

-Vinculum

Vinculum  Jun 14, 2022
 #6
avatar+280 
0

I think both of you guys messed up :)

 Jun 14, 2022
 #7
avatar+556 
+1

Wait @hipie what happened?

Vinculum  Jun 14, 2022

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