Answer:
- zeros: 1, 0.5±i√1.75
- h(x)/(x+7) = x^2 -17x +130 -840/(x+7)
Step-by-step explanation:
1.
There are a couple of ways that polynomial division by a binomial can be done. I will use different methods for the two problems so you can see them both.
In order to find all of the roots of the cubic, we can divide it by the known factor to obtain the quadratic factor. Then we can solve the quadratic in any of the usual ways.
Here, we will use synthetic division. The details are shown in the first attachment. That shows us that ...
f(x) = (x -1)(x^2 -x +2)
The quadratic has no integer factors, but we can find the zeros by completing the square.
x^2 -x +2 = 0
x^2 -x +1/4 = -2 +1/4 . . . . . . subtract 2; add 1/4 to complete the square
(x -1/2)^2 = -1.75
(x -1/2) = ±√-1.75 . . . . . . . . . take the square root
x = 0.5 ± i√1.75 . . . . . . . . . . complex zeros of f(x)
Of course, the value of x that makes the given factor zero is x=1 (which is the number at top left of the synthetic division).
The zeros of f(x) are 1, 0.5 -i√1.75, and 0.5 +i√1.75.
(The complex zeros can also be written as (1±i√7)/2.)
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2.
The long division of h(x) = x^3 -10x^2 +11x +70 by the binomial factor (x +7) is shown in the second attachment. As it shows, each quotient term is the quotient of the leading dividend term and the leading divisor term. The partial dividend that has a degree less than that of the divisor is called the remainder. Here, the remainder is -840. Just as the remainder from regular integer division can be added to the quotient as a fraction:
7 ÷ 2 = 3 r 1 = 3 1/2
so can the remainder from polynomial long division be added to the quotient as a fraction.
h(x)/(x +7) = x^2 -17x +130 -840/(x +7)
