For what value of π will the polynomial π(π₯) = 2π₯
3 β ππ₯
2 + 9π₯ β 11 have the same
remainder as when itβs divided by (π₯ β 2) and (π₯ + 3)?
P(x)=2x^3-mx^2+9x-11
A little synthetic division shows that the remainders are
-9m-92 and 23-4m
If they are equal, then m=-23
P(-3) = P(2) = 115
To find the value of π that will make the polynomial π(π₯) have the same remainder when divided by (π₯ β 2) and (π₯ + 3), we need to use the Remainder Theorem.
The Remainder Theorem states that if a polynomial P(x) is divided by (x - r), then the remainder is equal to P(r).
In this case, we need to find the remainder when π(π₯) is divided by (π₯ β 2) and (π₯ + 3).
For (π₯ β 2):
Substitute x = 2 into π(π₯):
π(2) = 2(2)^3 - π(2)^2 + 9(2) - 11
Simplifying:
π(2) = 16 - 4π + 18 - 11
π(2) = -4π + 23
For (π₯ + 3):
Substitute x = -3 into π(π₯):
π(-3) = 2(-3)^3 - π(-3)^2 + 9(-3) - 11
Simplifying:
π(-3) = -54 + 9π - 27 - 11
π(-3) = 9π - 92
Since we want the polynomial to have the same remainder, we equate the remainders from the two divisions:
-4π + 23 = 9π - 92
Now we can solve for π:
-4π - 9π = -92 - 23
-13π = -115
π = -115/-13
π = 8.85
Therefore, the value of π that will make the polynomial π(π₯) have the same remainder when divided by (π₯ - 2) and (π₯ + 3) is approximately 8.85.