Just a quick question about a questoin relating to factor theorem:
Find the value of k so that when is divided by x + k, the remainder is 3.
Thanks!
The division algorithm says we can find the quotient and remainder:
x^3+5x^2+6x+11 = q(x)(x+k)+3
Since these are equal, their evaluations are equal as well.
Let us evaluate this at x=-k.
Thus,
(-k)^3+5(-k)^2+6(-k)+11=q(-k)(0)+3
-k^3+5k-6k+11=3
k^3-5k+6k-8=0
Use the rational root theorem and see that k=4 is a solution. (In fact the only one).
Thus, k=4.
Hello, ty2391!
I used the Remainder Theorem:Find the value of k so that when x³ + 5x² + 6x + 11 is divided by x + k, the remainder is 3.
. . If f(a) = r, then f(x) ÷ (x - a) has a remainder of r.
We want to find a so that f(a) = 3.
If a is nonpositive, we can see that f(a) > 11.
. . Hence, a must be negative.
We find that: . f(-1) .= .9
. . . . . . . . . . .f(-2) .= .11
. . . . . . . . . . .f(-3) .= .13
. . . . . . . . . . .f(-4) .= . 3 . ← There!
Therefore: .k = 4
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Check
(x³ + 5x² + 6x + 11) ÷ (x + 4) . = . x² + x + 2, .remainder 3