Hi this is a question from Chapter 10: Polynomials of Terry Lee 3U.
"Given that the equation 2x^4+9x^3+6x^2-20x-24=0 has a triple root, factorise it completely."
Using the multiplicity theorem, I differentiated it twice and found the triple root as follows..
P'(x) = 8x^3+27x^2+12x-20
P''(x)=24x^2+54x+12
let P''(x)=0
24x^2+54x+12=0
(4x+1)(x+2)=0
x=-2 or -1/4
P(-2)
Hence, triple root at x=-2
and (x+2)^3 a factor
Now, the problem comes here..
The solution used "by inspection" method to factorise the polynomial, but since I'm not that good at "by inspection" hahaha I used sum and product of roots as follows..
(let a be the fourth root)
sum of roots: -2 -2 -2 + a = -9/2
Hence, a=3/2
and (x-3/2) is a factor
So, in summary P(x)=(x+2)^3(x-3/2)
but the answer is.. P(x)=(x+2)^3(2x-3)
so I seem to be missing a "2" somewhere :I
So are there any loopholes with using sum and product to factorise a polynomial??
This was such a long post but it'd be great if anyone can help, thank you!!
"Given that the equation 2x^4+9x^3+6x^2-20x-24=0 has a triple root, factorise it completely."
Using the multiplicity theorem, I differentiated it twice and found the triple root as follows..
P'(x) = 8x^3+27x^2+12x-20
P''(x)=24x^2+54x+12
let P''(x)=0
24x^2+54x+12=0
(4x+1)(x+2)=0
x=-2 or -1/4
P(-2)
Hence, triple root at x=-2
and (x+2)^3 a factor
Now, the problem comes here..
The solution used "by inspection" method to factorise the polynomial, but since I'm not that good at "by inspection" hahaha I used sum and product of roots as follows..
(let a be the fourth root)
sum of roots: -2 -2 -2 + a = -9/2
Hence, a=3/2
and (x-3/2) is a factor
So, in summary P(x)=(x+2)^3(x-3/2)
but the answer is.. P(x)=(x+2)^3(2x-3)
so I seem to be missing a "2" somewhere :I
So are there any loopholes with using sum and product to factorise a polynomial??
This was such a long post but it'd be great if anyone can help, thank you!!
