Question 1159546: The function f is given by f(x)=x^3+2x^2+ax-8 where a is constant.when f(x) is divided by (x-2) the remainder is -6.
Show that (x+1) is a factor of f(x)
Found 3 solutions by ikleyn, MathLover1, MathTherapy: Answer by ikleyn(52777) (Show Source):
You can put this solution on YOUR website! .
According to the Remainder theorem, the fact that the remainder is -6, when f(x) is divided by (x-2), means that f(2) = -6.
In other words,
2^3 + 2*2^2 + a*2 - 8 = -6.
It implies
2a = -6 - 2^3 - 2*2^2 + 8 = -14.
Hence, a = -14/2 = -7.
Thus the polynomial f(x) is
f(x) = x^3 + 2x^2 - 7x - 8. (1)
Now, let us check that f(-1) = 0.
For it, substitute x= -1 into the polynomial (1). You will get
f(-1) = (-1)^3 + 2*(-1)^2 - 7*(-1) - 8 = -1 + 2 + 7 - 8 = 0.
Now apply the Remainder theorem again.
It says that if f(-1) = 0, then f(x) is divisible by the binomial (x-(-1)) = (x+1).
It is EXACTLY what has to be proved.
The proof is completed.
The problem is Solved.
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Theorem (the remainder theorem)
1. The remainder of division the polynomial by the binomial is equal to the value of the polynomial.
2. The binomial divides the polynomial if and only if the value of is the root of the polynomial , i.e. .
3. The binomial factors the polynomial if and only if the value of is the root of the polynomial , i.e. .
See the lessons
- Divisibility of polynomial f(x) by binomial x-a and the Remainder theorem
- Solved problems on the Remainder thoerem
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Divisibility of polynomial f(x) by binomial (x-a). The Remainder theorem".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.
Answer by MathLover1(20849) (Show Source): Answer by MathTherapy(10551) (Show Source):
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