find the value of k such that when 2x^3+9x^2+kx-15 is divided by x+5, the remainder is 0.
In google type :
2x^3+9x^2+kx-15 is divided by x+5, the remainder is 0
When you see list of results click on
Y a h o o! Canada Answers - Polynomial Division Problem?
You will see solution with explanation.
in fact , u can substitude -5 into the x variable,and the write it like this:
2(-5)^3+9(-5)^2+k(-5)-15=0
then plz find k
6x3 + 29x2 + 2x – 90 is divided by x + 3
To find the value of k such that the polynomial 2x^3 + 9x^2 + kx - 15 is divided by x + 5 with a remainder of 0, we will use the Remainder Theorem.
The Remainder Theorem states that if a polynomial P(x) is divided by x - a, then the remainder is P(a). In this case, we are dividing by x + 5, so our "a" in the theorem is -5.
So, we need to evaluate the polynomial 2x^3 + 9x^2 + kx - 15 at x = -5 and set the result to 0, since the remainder should be zero.
Substituting x = -5 into the polynomial, we get:
2(-5)^3 + 9(-5)^2 + k(-5) - 15 = 0
Simplifying, we have:
-250 + 225k - 5k - 15 = 0
Combining like terms, we get:
-250 + 220k - 15 = 0
Simplifying further, we have:
220k - 265 = 0
To isolate the variable, we can add 265 to both sides:
220k = 265
Finally, divide both sides by 220 to find the value of k:
k = 265/220
Simplifying the division, we have:
k = 1.20454545
Therefore, the value of k that makes the remainder zero when dividing 2x^3 + 9x^2 + kx - 15 by x + 5 is k = 1.20454545.