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# 1. Let . For what value of b is f(1)=1?

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1.  Let \(f(x)=3x^4-7x^3+2x^2-bx+1\). For what value of b is f(1)=1?

2.  Find the constant c such that (x^2-4x+3)(x+5) - (x^2+4x-5)(x-c)=0 for all x.

Nov 21, 2019

#1
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Is this your homework or for a test? Because it sure looks like it. If you confirm its not I will help you!   Nov 21, 2019
#2
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its not if was just looking through different websites and wanted to learn how to do this.

Guest Nov 21, 2019
#4
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These are Alcumus problems, so yeah, this is totally homework.

Guest Nov 21, 2019
#5
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1.

f(1)  =   3 - 7 + 2 - b + 1  =  1

So

-1 - b  = 1     add 1 to both sides

-b = 2

b = -2   Nov 21, 2019
#6
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2.

(x^2-4x+3)(x+5) - (x^2+4x-5)(x-c) = 0    simplify

(x^3 - 4x^2 + 3x) + (5x^2 - 20x + 15)  - (x^3 + 4x^2 - 5x - cx^2 - 4cx + 5c)

(x^2 - 17x + 15)  - 4x^2 + 5x + cx^2 + 4cx - 5c  = 0

(c - 3)x^2  + (4c - 12) x + (15 - 5c)  =   0

(c - 3)x^2 + 4(c - 3)x - 5(c - 3)  =  0

(c - 3) (x^2 + 4 - 5)  =  0

c = 3       will make this true   Nov 21, 2019