Problem 7
<h3>Answer: Choice A) 7</h3>--------------------------------------
Work Shown:
p(x) = x^4+x^3-kx-x+6
If (x-2) is a factor of p(x), then x = 2 is a root of p(x). This is a special case of the remainder theorem.
This means p(2) = 0
Replace every x with 2 and solve for k
p(x) = x^4+x^3-kx^2-x+6
p(2) = 2^4+2^3-k(2)^2-2+6
p(2) = 16+8-4k-2+6
p(2) = 28-4k
0 = 28-4k .... replace p(2) with 0
4k = 28
k = 28/4
k = 7
The polynomial
p(x) = x^4+x^3-7x^2-x+6
has the factor (x-2)
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Problem 8
<h3>Answer: D) 40</h3>--------------------------------------
Work Shown:
Check out the diagram below to see the synthetic division table. We're after the remainder which is highlighted in yellow.
An alternative is to use direct substitution
p(x) = 2x^4+4x^3-x^2-5
p(-3) = 2(-3)^4+4(-3)^3-(-3)^2-5 ... replace every x with -3
p(-3) = 2(81)+4(-27)-9-5
p(-3) = 162-108-9-5
p(-3) = 40
This helps confirm we got the right remainder and the right answer.
