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If the product (3x^2 - 5x + 4)(7 - 2x) can be written in the form ax^3 + bx^2 + cx + d, where a,b,c,d are real numbers, then find 8a + 4b + 2c + d.

 Apr 20, 2021
 #1
avatar+118505 
+1

(3x^2  -  5x  +  4)    ( 7  -  2x)

 

7 ( 3x^2  -  5x  + 4)  =    21x^2  -35x  +  28

 

-2x ( 3x^2  -5x  + 4)  = -6x^3 + 10x^2  - 8x

 

Adding like terms  we  get

 

-6x^2  +  31x^2  - 43x + 28

 

a = -6       8a =  -48

b = 31      4b = 124

c =  -43    2c  = -86

d =  28

 

 

cool cool cool        

 Apr 20, 2021
 #2
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+1

The answer is 18, actually.

\(8a+4b+2c+d=(3 \cdot (2)^2 - 5 \cdot (2) + 4)(7 - 2 \cdot (2)) = 6 \cdot 3 = \boxed{18}\)

 Apr 20, 2021

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