P(x) = x^4 - 8x^2 + 12 ???
P(x) is a monic polynomial of the fourth degree. When P(x) is divided by x+1 and x-2 the remainders are 5 and -4 respectively. Given that P(x) is an even function ie. one where P(x)=P(-x)
a. Express it in the form p0+p1x^1+p2x^2+p3x^3+p4x^4
Well what I instantly thought was equating both by subbing in the zeroes, but what equation would I substitute it in?
Maybe P(x)=A(x)Q(x)+R(x)?
well what I know is P(-1) = 5
and P(2) = -4
and then at one point, I would have to use the condition P(x)=P(-x)
Any help please? thanks.
P(x) = x^4 - 8x^2 + 12 ???
1-on-1 Maths Tutoring(IB & HSC): Epping, Beecroft, Eastwood, Carlingford & Beyond
IB: Maths Studies, Maths SL & Maths HL; HSC: 2U, 3U & 4U
Highly Qualified & Highly Experienced. Estimated ATAR > 9.995
There are IB Maths Tutors and there are IB Maths Tutors.
2013 HSC - Standard English | Mathematics Extension 1 | Mathematics Extension 2 | Physics | Chemistry
One more thing,
when the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?
I know that R(1) = 2, which is the answer, but what could be a nice proof?
Once you've used the fact that the polynomial is monic, so the coefficient of the x^4 is 1, and that it is an even function, so there is no odd power of x, you can write p(x)=x^4+bx^2+d. Now the polynomial has only two unknowns, b and d, and you have two pieces of information p(-1)=5 and p(2)=-4. When you use this you end up getting the two simulataneous equations b+d=4 and 4b+d=-20. Find the value of b & d and bob's your uncle
2nd Year Actuarial Studies/Applied Finance
thanks
There are currently 1 users browsing this thread. (0 members and 1 guests)
Bookmarks