Question

Question no 6 answer

Answer 1

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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Comment on the attached graph

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