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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, and d are real numbers, then find 8a + 4b + 2c + d.

 


Thanks dudes and dudettes!

AnonymousConfusedGuy  Nov 30, 2017
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 #1
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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, and d are real numbers, then find 8a + 4b + 2c + d.

 

(3x^2 - 5x + 4)(7 - 2x) 

=(7 - 2x) (3x^2 - 5x + 4)     It is easier if you put the one with the fewest terms first.

 

Now just take the first term in the first bracket and multiply it by the second bracket ... 

     then take the 2nd term and multiply it by the second bracket ... 

           etc until you run out of terms in the first bracket. 

 

\(=(7 - 2x) (3x^2 - 5x + 4)\\ =   7(3x^2 - 5x + 4)  \quad   - 2x (3x^2 - 5x + 4)\\ = 21x^2-35x+28 \quad -6x^3+10x^2-8x\\ =-6x^3+21x^2+10x^2-35x-8x+28 \\ =-6x^3+31x^2-43x+28 \\ so\\ a=-6,\quad b=31, \quad c=-43, \quad and \quad d=28 \)

 

You can work out what 8a + 4b + 2c + d. equals :)

 

If you have questions then ask :)

Melody  Nov 30, 2017
 #2
avatar+257 
+2

Thanks! :)

AnonymousConfusedGuy  Nov 30, 2017

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