polynomials (1 Viewer)

maths lover

Member
Joined
Feb 22, 2011
Messages
292
Gender
Male
HSC
2012
p of x=x^3-8x^2+9x+18 is divisible by (x-3) and (2x+1). by considering the leading term and the constant term, express the polynomial in terms of three linear factors. and hence solve p(X) is less that or equal to 0.

i know how to do the question since i got the one before it but for some reason my answer has small difference in the solving part.
thanxs any help is appreciative
 

maths lover

Member
Joined
Feb 22, 2011
Messages
292
Gender
Male
HSC
2012
that is not the answer in the textbook. the book says x<=-2 or -1/2<=x<=3
 

maths lover

Member
Joined
Feb 22, 2011
Messages
292
Gender
Male
HSC
2012
im really sorry but i wrote down the wrong polynomial it is p of x=2x^3-x^2-13x-6
 

funnytomato

Active Member
Joined
Jan 21, 2011
Messages
848
Gender
Male
HSC
2010
im really sorry but i wrote down the wrong polynomial it is p of x=2x^3-x^2-13x-6
okay

so let the other factor be ax+b
then P(x)= 2x^3-x^2-13x-6 = (x-3)(2x+1)(ax+b)

equating coefficient of x^3 , 2=2a, a=1
... of x^0 , -6 = -3b , b=2
P(x)= 2x^3-x^2-13x-6 = (x-3)(2x+1)(x+2) , which has roots x=3, x=-1/2 and x=-2

after drawing a simple graph(a cubic with +ve leading coefficient) , for P(x) <= 0
x<= -2 or -1/2<=x<=3
 

maths lover

Member
Joined
Feb 22, 2011
Messages
292
Gender
Male
HSC
2012
sorry to ask, but could we say through inspection obviously a=1 and b=2.
 

funnytomato

Active Member
Joined
Jan 21, 2011
Messages
848
Gender
Male
HSC
2010
p of x=x^3-8x^2+9x+18 is divisible by (x-3) and (2x+1). by considering the leading term and the constant term, express the polynomial in terms of three linear factors. and hence solve p(X) is less that or equal to 0.

i know how to do the question since i got the one before it but for some reason my answer has small difference in the solving part.
thanxs any help is appreciative
sorry to ask, but could we say through inspection obviously a=1 and b=2.
the question is explicitly asking you to "consider" the coefficients , I'd write them down to be apparent
otherwise , you could state the values of a and b by inspection
or use sum of roots since you only need the roots to graph it and then solve the inequality
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top