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You are here: Experts > Science > Mathematics > Advanced Math > remainder theorem
Expert: Sherry Wallin - 11/2/2009
Question
a cubic polynomial gives remainders (5x+4)and (12x-1) when divided by x^2-x+2 and x^2+x-1 respectively. find the polynomial.
2. when polynomial P(x) is divided by (x-b)it leaves a remainder of b^3 and (x-a^3) the remainder is a^3. find the remainder when p(x) is divided by (x-a)(x-b).
function
3. fx= (x+10)/(x-8), x not = 8, x>/= 0. determined the range.
4. sktech the graph of fx=1/a (log base a (ax-1)), 0<a<1. hence find x:f(x)>1/a
what is the differences between y=|f(x)| and y=f|(x)|
Answer
Belle~
There are a number of ways to approach these problems. I think your teacher wants you to use the remainder theorem which says that if you divide a polynomial by x - c then the remainder is f(c). An easy example to see this is if I tell you that the polynomial is x^+2x + 2 and I divide it by (x+1) and tell you that it's remainder is 1 then f(-1) = 1 let's check it out:
f(-1) = (-1)^2+2(-1) + 2 = 1-2+2 = 1. Here in your first problem your divisor is not a 'linear factor', i.e. your divisor is not of the form x - c, however since you are dividing a 3rd degree polynomial by a 2nd degree polynomial you will get a linear quotient which is a divisor also. What I mean by this is whatever remainder you get upon division, it will be the same for 'both' divisors. For example if I was to divide 7 by 2 I would get 3 remainder 1, or I could divide 7 by 3 and I would get 2 remainder 1. Both 2 and 3 were used as divisors and they both generated a remainder of 1. so when you divide a 3rd degree polynomial by a 2nd degree polynomial your answer will be a linear (first degree) polynomial plus a remainder. A standard 3rd degree polynomial 'looks' like:
ax^3+bx^2+cx+d. What I did was I used long division to divide the standard polynomial by each of x^2-x+2 and x^2+x-1 and got remainders of (-a+b+c)x +-2a-2b+d and (2a-b+c)x + -a+b+d respectively.
You can then compare coefficients with each of the remainders so comparing (-a+b+c)x +-2a-2b+d with 5x + 4 tells me that -a+b+c = 5 and -2a-2b+d = 4 and likewise with the second division we have
2a-b+c = 12 and -a+b+d = -1. You have a system of equations to solve. Notice two of the equations only have a,b,c in them and two have only a,b,d in them. Solve the systems and you will find that a = 1, b = -2, c = 8 and d =2, i.e your cubic polynomial is x^3-2x^2+8x+2.
I can only help you on one more problem so let's look at 3:
This is a rational equation which means there will be vertical asymptote(s). In this case
fx = (x+10)/(x-8), and there is only one factor to make the denominator zero so you will have a vertical asymptote at x = 8. The graph will not cross at x = 8, that is why you were given x not = 8,and you were given x>/= 0 so you only have to look at what happens to the graph in the first and fourth quadrants and to the right of the y axis .x+10 will always be positive for x positive but until x = 8, x - 8 will be negative. So the range is a positive number over a negative number until x > 8 which means the range will be in the 4th quadrant where y is f(x) is negative. How negative you might ask. At x = 0 you have 10/ -8 =-5/4 and when x = 7 you have 17/-1. Try getting as close as possible to 8 without touching 8, let's try 7.9999 for example, this gives us 17.9999/-.0001 = -179,999. It should be obvious that y will get as small as you like on the left of x = 8. How about on the right of 8. Reason it out the same way, look at x = 8.1 and maybe 90. When x is 8.1 you get 18.2/.1 = 181, but how about when x = .0001? You will get 10.0001/.0001 = 100,001. It should be a apparent that on the right side of x = 8 that y gets very large. And finally look at x = 90. then you get 100/82 = about 1.2. You could try x = a million and then you'd have 1,000,010/999992 = about 1.000018. So your range is -5/4 at x = 0 to negative infinity approaching x = 8 from the left and then approaching x = 8 from the right y goes to infinity and moving left to right from the right side of x = 8 you get really close to 1 so y = 1 is a horizontal asymptote. i.e., on the right hand side of x = 8 the graph is always above 1 and as high as infinity and on the left side of x = 8 the graph is -5/4 and as low as negative infinity. There are no y -values between (-5/4,1) hence the range is all real numbers except those in (-5/4,1).
Good luck,
Math Prof
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