if the polynomial f(x)=ax^{3}+(a+b)x^{2}+(a+2b)x+1is divisible by x+1, find the quotient, leaving its coefficients in term of a only. Prove that in this case, the equation f(x) has only one real root if a^{2}-6a+1

Question

if the polynomial f(x)=ax^{3}+(a+b)x^{2}+(a+2b)x+1is divisible by x+1, find the quotient, leaving its coefficients in term of a only. Prove that in this case, the equation f(x) has only one real root if a^{2}-6a+1<0.Hence, show that 3-2sqrt{2}<a<3+2sqrt{2}

anonymous · Answer

$$f(x)=ax^{3}+(a+b)x^{2}+(a+2b)x+1$$divisible by $$x+1$$means$$f(-1)=0$$$$f(-1)=a(-1)^{3}+(a+b)(-1)^{2}+(a+2b)(-1)+1=0$$$$b=1-a$$therefore$$f(x)=ax^{3}+(a+1-a)x^{2}+(a+2(1-a))x+1$$$$f(x)=ax^{3}+x^{2}+(2-a)x+1$$let$$f(x)=(x+1)(ax^{2}+Ax+1)$$compare coefficient of x,$$2-a=1+A$$$$A=1-a$$therefore,$$f(x)=(x+1)(ax^{2}+(1-a)x+1)$$when f(x)=0,$$0=(x+1)(ax^{2}+(1-a)x+1)$$then i stuck here....

myininaya · Answer

neoh i'm so becoming your fan
no one ever shows their attempt

anonymous · Answer

thx...

myininaya · Answer

i got b=-1-a

myininaya · Answer

did i make a mistake there or did you?

anonymous · Answer

oh no... i make a mistake... but if using the same method i also stuck at same place...

dumbcow · Answer

its b =1-a
its a "+1" at the end

myininaya · Answer

you are good neoh
i made the mistake

dumbcow · Answer

looks good
last step, set equal to zero for x=-1
$$ax^{2} +(1-a)x+1 = 0$$
$$a(-1)^{2} +(1-a)(-1)+1 = 0$$
solve for a

dumbcow · Answer

wait nevermind, scratch that

dumbcow · Answer

oh you need to show there is only 1 real root.
use discriminant < 0

anonymous · Answer

i already factorize f(x) by x+1, so the quotient can substitute the same zero again or not??

myininaya · Answer

$$(1-a)^2-4a<0$$

anonymous · Answer

$$(1−a)^{2}-4(a)(1)<0$$$$1−2a+a^{2}-4a<0$$$$a^{2}-6a+1<0$$

dumbcow · Answer

discriminate : b^2 - 4ac
$$\rightarrow (1-a)^{2} -4a < 0$$
$$a^{2}-6a+1 < 0$$

myininaya · Answer

lol yeah!

anonymous · Answer

dumbcow is right, just show the discriminant being less than 0.   discriminant of
$$ax^2+(1-a)x+1 $$

anonymous · Answer

myininaya - I am hurt.. I always do :P

myininaya · Answer

what?

dumbcow · Answer

haha good job neoh...now just use quadratic formula

anonymous · Answer

I always reply with the procedure.. not just the answer :D Jus Kidding

anonymous · Answer

$$(-b \pm \sqrt{b^{2}-4ac})\div 2a$$

myininaya · Answer

neat problem

myininaya · Answer

kind of long but neat

myininaya · Answer

i'm going to sleep later guys

anonymous · Answer

night

anonymous · Answer

$$(6 \pm \sqrt{(-6)^{2}-4(1)(1)})\div2(1)$$$$3 \pm 2\sqrt{2}$$

anonymous · Answer

have two roots... the question say 1 root.... how to prove that??

anonymous · Answer

to complete the problem you need to check the intervals:
$$(-\infty, r_1], (r_1,r_2), [r_2, \infty)$$
So that if:
$$r_1<a<r_2$$
there is only one real root to your equation.

anonymous · Answer

sry, where:
$$r_1 = 3-2\sqrt{2}, r_2 = 3+2\sqrt{2}$$

anonymous · Answer

and try to keep the perspective straight, these roots we found arent for the polynomial x, they are roots of a.

anonymous · Answer

ok...

anonymous · Answer

If there is only 1 real root then the root must be -1.  That means that :
$$ax^2+(1−a)x+1$$
has no real solution.  Using the quadratic formula:
$$x = \frac{-(1-a) \pm \sqrt{(1-a)^2-4a}}{2a}$$ must not be real thus, the thing under the radical must be negative.  Thus
$$4a > (1-a)^2 = 1-2a+a^2$$
$$0>1-6a+a^2$$
as required

anonymous · Answer

ok...