TERRY LEE QUESTION - factorising using sum and product of roots (1 Viewer)

sazkim · Oct 6, 2016

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!!

InteGrand · Oct 6, 2016

sazkim said:
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!!

$\noindent Your answer would be right if (and only if) the polynomial was monic (had leading coefficient $1$). But the leading coefficient is $2$, so you need to put a $2$ in front of your answer. You can then obtain Terry Lee's answer by multiplying the $2$ into the $\left(x - \frac{3}{2}\right)$.$

$\noindent Remember, in general, the factorisation of a degree $n$ polynomial $p(x)$ is $A \left(x-\alpha_1\right)\ldots \left(x - \alpha_{n}\right)$, where $\alpha_1,\ldots, \alpha_{n}$ are the roots and $A$ is the leading coefficient of $p(x)$.$

tsoliman1 · Oct 9, 2016

InteGrand said:
$\noindent Your answer would be right if (and only if) the polynomial was monic (had leading coefficient $1$). But the leading coefficient is $2$, so you need to put a $2$ in front of your answer. You can then obtain Terry Lee's answer by multiplying the $2$ into the $\left(x - \frac{3}{2}\right)$.$

$\noindent Remember, in general, the factorisation of a degree $n$ polynomial $p(x)$ is $A \left(x-\alpha_1\right)\ldots \left(x - \alpha_{n}\right)$, where $\alpha_1,\ldots, \alpha_{n}$ are the roots and $A$ is the leading coefficient of $p(x)$.$

Can't u also do long division with (x-2)^3 after expanding to get the other factor?

InteGrand · Oct 9, 2016

tsoliman1 said:
Can't u also do long division with (x-2)^3 after expanding to get the other factor?

Yeah, you can.

Hughmaster · Oct 11, 2016

Long Division is an absolute trek though - I would strongly advise against it

TERRY LEE QUESTION - factorising using sum and product of roots (1 Viewer)

sazkim

New Member

InteGrand

Well-Known Member

tsoliman1

Member

InteGrand

Well-Known Member

Hughmaster

Hugh Entwistle

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