I divided the x+3 by the other polynomial.
What I did was factor the denominator to (x+3)(x-1)
The x+3 cancels with the one in the denominator
I was left with 1/(x-1)
f(0) would give me -1 and f(4) would give me 1/3 so I have none of the above.
If I didn't understand what you meant above... then maybe we are dividing (x+3) into the other polynomial. If that is the case.. we are left with x -1
f(0)= -1 and f(4)= 3 so the answer is 2... I think this is more likely what you were supposed to do.
to divide a function f(x+3) by (x^2+2x-3), we see the quotient is Q(x) and remainder is (x+2). If f(0)=a and f(4)=b, what is the value of (a+b)
a)2
b)4
c)8
d)10
e) none of the above
3 answers
You said:
f(x+3) / (x^2+2x-3) = Q(x) + (x+2)/(x^2+2x-3)
f(x+3) = (x^2+2x-3)Q(x) + x+2
we want f(0), so our x+3 = 0 or x = -3
f(0) = (9 - 6 - 3)Q(-3) + -3+2 = a
a = -1
we also want f(4), so our x+3 = 4, which means x = 1
f(4) = (1 + 2 - 3)Q(1) + 1+2 = b
b = 3
so
a+b = -1+3 = 2
f(x+3) / (x^2+2x-3) = Q(x) + (x+2)/(x^2+2x-3)
f(x+3) = (x^2+2x-3)Q(x) + x+2
we want f(0), so our x+3 = 0 or x = -3
f(0) = (9 - 6 - 3)Q(-3) + -3+2 = a
a = -1
we also want f(4), so our x+3 = 4, which means x = 1
f(4) = (1 + 2 - 3)Q(1) + 1+2 = b
b = 3
so
a+b = -1+3 = 2
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