Polynomial Question (1 Viewer)

[CanU_Not]

Could someone please solve this (and explain how to approach these sort of questions):

Suppose a polynomial is given by P(x)= (x-a)(x-b)(x-c)-(b+c)(c+a)(a+b).
a.) Show that x=a+b+c is a zero of P(x)
B.) Hence, or otherwise, factorise the polynomial P(x)= (x-2)(x+3)(x+1)-4.

Answer for a wasn't given as it is a proof.
Answer for b= (x+2)(x-sqrt(5)) (x+sqrt(5))

EDIT: THANKS FOR ALL THE REPLIES )

Another polynomial question :'(
"By solving the line y=mx+b simultaneously with the cubic y= x^3-6x^2-2x+1 and insisting that there be a triple root, find the point of inflexion of the cubic without using calculus"

Thanks in advance

 

[lpodtouch]

For part a)
x-(a+b+c) is the root
P(a+b+c) = (a+b+c-a)(a+b+c-b)(a+b+c-c) - (b+c)(c+a)(a+b)
= (b+c)(a+c)(a+b) - (b+c)(c+a)(a+b)
= 0

For part b)
Factorising P(x) = (x-2)(x+3)(x+1)-4
Using part a), let a=2, b=-3, c=-1 (corresponding numbers from part a)
As x=a+b+c is a zero for P(x)
Therefore, x= 2-3-1 = -2
x+2 is a zero for P(x) = (x-2)(x+3)(x+1)-4
Multiply the constant numbers of the factorised P(x) = -2 x 3 x 1 and then minus 4
= -10
As you know that the constant from the x+2 zero is 2, therefore the other roots must be -sqrt5 and sqrt 5

Ergo, the polynomial factorised is:
P(x)= (x+2)(x-sqrt(5))(x+sqrt(5))

Sorry for the terrible formatting, but hope it helped ;P

 

[integral95]

For part a)
x-(a+b+c) is the root
P(a+b+c) = (a+b+c-a)(a+b+c-b)(a+b+c-c) - (b+c)(c+a)(a+b)
= (b+c)(a+c)(a+b) - (b+c)(c+a)(a+b)
= 0

For part b)
Factorising P(x) = (x-2)(x+3)(x+1)-4
Using part a), let a=2, b=-3, c=1 (corresponding numbers from part a)
As x=a+b+c is a zero for P(x)
Therefore, x= 2-3-1 = -2
x+2 is a zero for P(x) = (x-2)(x+3)(x+1)-4
Multiply the constant numbers of the factorised P(x) = -2 x 3 x 1 and then minus 4
= -10
As you know that the constant from the x+2 zero is 2, therefore the other roots must be -sqrt5 and sqrt 5

Ergo, the polynomial factorised is:
P(x)= (x+2)(x-sqrt(5))(x+sqrt(5))

Sorry for the terrible formatting, but hope it helped ;P

How can you tell.. I know you used product of roots and that the product of the other 2 roots is 5..it could be like 1 and 5 for all we know

 

[QZP]

For part a)
x-(a+b+c) is the root
P(a+b+c) = (a+b+c-a)(a+b+c-b)(a+b+c-c) - (b+c)(c+a)(a+b)
= (b+c)(a+c)(a+b) - (b+c)(c+a)(a+b)
= 0

For part b)
Factorising P(x) = (x-2)(x+3)(x+1)-4
Using part a), let a=2, b=-3, c=-1 (corresponding numbers from part a)
As x=a+b+c is a zero for P(x)
Therefore, x= 2-3-1 = -2
x+2 is a zero for P(x) = (x-2)(x+3)(x+1)-4
Multiply the constant numbers of the factorised P(x) = -2 x 3 x 1 and then minus 4
= -10
As you know that the constant from the x+2 zero is 2, therefore the other roots must be -sqrt5 and sqrt 5

Ergo, the polynomial factorised is:
P(x)= (x+2)(x-sqrt(5))(x+sqrt(5))

Sorry for the terrible formatting, but hope it helped ;P

What did you do here? Sorry I don't follow

 

[lpodtouch]

How can you tell.. I know you used product of roots and that the product of the other 2 roots is 5..it could be like 1 and 5 for all we know

You use factor theorem to ensure whether it is either 1 or -1
Considering how P(1) = (1-2)(1+3)(1+1) -4
= -12
and P(-1) = (-1-2)(-1-3)(-1+1) - 4
= -4

You use the sqrts of 5 instead

I just skipped that step because it was quicker to deduce. In future, utilise factor theorem to ensure accuracy.

 

[lpodtouch]

What did you do here? Sorry I don't follow

Considering how it is a monic cubic polynomial (upon finding the leading term), you can quicken the process of finding the roots through looking for the factors of the constant term (of the polynomial)
In this case it is -10 upon multiplication

Considering how we have found x+2 is a zero for P(x)=(x-2)(x+3)(x+1)-4, we require a -5 from the other two zeroes.
Factors to consider are: 1, -1, 5, -5, sqrt5 and sqrt-5

Utilise factor theorem through the original polynomial given P(x)=(x-2)(x+3)(x+1)-4
If P(x)= 0 then you have figured out your roots

Sorry for being brief - my teacher taught me this method to quicken the somewhat laborious process of factorising polynomials :/

 

[QZP]

I don't like how you have +-sqrt 5 as zeros to consider, because that thereom only applies to integer zeros (if I recall correctly) - it feels like you only listed it because you saw the answer. For safety just expand the polynomial and factorise (x+2). Doesn't take that long anyway

 

[lpodtouch]

I don't like how you have +-sqrt 5 as zeros to consider, because that thereom only applies to integer zeros (if I recall correctly) - it feels like you only listed it because you saw the answer. For safety just expand the polynomial and factorise (x+2). Doesn't take that long anyway

Yeah factor theorem is applicable to integer zeroes (however upon substituting sqrt5 and -sqrt5 = 0)

It would preferable to just expand the polynomial and long divide to receive P(x) = (x+2)(x^2-5) then factorise accordingly.

 

[CanU_Not]

Another polynomial question :'(
"By solving the line y=mx+b simultaneously with the cubic y= x^3-6x^2-2x+1 and insisting that there be a triple root, find the point of inflexion of the cubic without using calculus"

Thanks in advance

 

[aDimitri]

if you doubt the +-sqrt(5) you can expand and check sum of roots (you will see it is -2) and since -2 is the known root the other roots are opposite in sign and equal in magnitude.

 

[dunjaaa]

Think about this graphically. When we find the points of intersection, we equate the y's. The new cubic polynomial formed is x^3-6x^2-(m+2)x+1-b=0. We are given that there exists a triple root. Hence, this implies that the line y=mx+b must of intersected the original polynomial at its point of inflexion via subtraction of ordinates. Thus, we must solve for x on the new cubic polynomial. Since there exists a triple root, we must transform the new cubic polynomial into the form (x+a)^3 (monic) for some real number 'a'. Expanding (x+a)^3, we obtain x^3+3ax^2+3(a^2)x+a^3. Equating the x^2 terms, we get 3a=-6, a=-2. Note: The line can also be found as well, by equating both the x and constant terms. You should get y=-14x+9. Therefore the new cubic polynomial can be transformed into (x-2)^3. This means that point of inflexion must of existed at x=2. Substituting into the original cubic to find the y-ordinate, we obtain y=-19. Therefore POI exists at (2,-19).