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.
+22 Is this your homework or for a test? Because it sure looks like it. If you confirm its not I will help you!
+129899 1.
f(1) = 3 - 7 + 2 - b + 1 = 1
So
-1 - b = 1 add 1 to both sides
-b = 2
b = -2
+129899 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
