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# polynomial

0
236
7

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

#1
+580
+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

Jun 14, 2022

#1
+580
+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

Vinculum Jun 14, 2022
#3
+1

Think again, Vinculum

Guest Jun 14, 2022
#4
+580
+1

Edit : 3+7+14+50

its 74

Vinculum  Jun 14, 2022
#2
+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
+580
+1

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

-Vinculum

Vinculum  Jun 14, 2022
#6
+344
0

I think both of you guys messed up :)

Jun 14, 2022
#7
+580
+1

Wait @hipie what happened?

Vinculum  Jun 14, 2022