Statements showing a relationship between numbers that are not necessarily equal using the symbols >, <, ≥, ≤, or ≠ are ca
1 answer:
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Answer:
- a = 1
- b = -28
- P(x) = (x +2)(x -3)(x^2 +3x +10)
Step-by-step explanation:
a) Since x+2 is a factor, we know P(-2) = 0.
P(x) = (((x +2)x +a)x +b)x -60
Then the value of P(-2) is ...
P(-2) = 0 = (((0)(-2) +a)(-2) +b)(-2) -60 = (-2a +b)(-2) -60 = 4a -2b -60
We know the remainder from division by (x+3) is 60, so
P(-3) = 60 = (((-3+2)(-3) +a)(-3) +b)(-3) -60 = ((3+a)(-3) +b)(-3) -60
= (-9 -3a +b)(-3) -60 = 27 +9a -3b -60
93 = 9a -3b
These two equations can be put into standard form:
2a -b = 30
3a -b = 31
Then we have the solution ...
a = 1 . . . . . (by subtracting the first equation from the second)
-28 = b . . . by substituting into the first equation
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b) To show that (x-3) is a factor we need to evaluate P(3).
P(3) = (((3 +2)(3) +1)(3) -28)(3) -60 = (48 -28)(3) -60 = 0
The function value is 0, so (x -3) is a factor.
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c) We want to find Q(x) = x^2 +cx +d such that ...
(x +2)(x -3)Q(x) = P(x)
(x^2 -x -6)(x^2 +cx +d) = x^4 +2x^3 +x^2 -28x -60
x^4 +(c-1)x^3 +(-6-c+d)x^2 +(-6c-d)x -6d = x^4 +2x^3 +x^2 -28x -60
This gives rise to the equations ...
c -1 = 2 ⇒ c = 3
-6d = -60 ⇒ d = 10
Then P(x) can be factored as ...
P(x) = (x +2)(x -3)(x^2 +3x +10)
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<em>Comment on the attached graph</em>
I like to use a graphing calculator to find real roots of higher-degree polynomials. This graph shows the real zeros to be -2 and +3, so we know that (x +2) and (x -3) are factors. The green curve is P(x) with those factors divided out, so is a graph of Q(x). The vertex of that graph tells us that ...
Q(x) = (x +1.5)^2 +7.75 = x^2 +3x +10
