polynomials question (1 Viewer)

[namburger]

Find all the pairs of values of the integers a and b for which the polynomial
P(x)= (ax+b)^2 −x is exactly divisible by both (x − 1) and (x − 4).

Help

 

[shaon0]

Let x=1.
P(1)=(a+b)^2 -1....1
Let x=4
P(4)=(4a+b)^2 -4....2
Solve simultaneously
Sorry i am doing Chemistry at the moment

 

[lolokay]

Find all the pairs of values of the integers a and b for which the polynomial
P(x)= (ax+b)^2 −x is exactly divisible by both (x − 1) and (x − 4).

Help

just expand and equate coefficients

 

[shaon0]

yea.....
250th post!

 

[lolokay]

and what a fine post it is

 

[lyounamu]

Find all the pairs of values of the integers a and b for which the polynomial
P(x)= (ax+b)^2 −x is exactly divisible by both (x − 1) and (x − 4).

Help

Since it is devisible by (x-1) and (x-4), substituting x=1 or x=4 will yield P(x)=0
i.e. P(1) = (a . 1 + b)^2 - 1 = 0 ...(1)
And P(4) = (a . 4 +b)^2 -4 = 0...(2)

From the equation (1), a^2 + 2ab + b^2 - 1 = 0
(1) x 4: 4a ^2 + 8ab + 4b^2 - 4 = 0...(3)
From the equation (2), 16a^2 + 8ab + b^@ - 4 = 0
(2) - (3):12a^2 - 3b^2 = 0
4a^2 = b^2
b = 2a or b = -2a
Substitute 2a for the value of b in the equation (1)
i.e. a^2 + 2a . 2a + (2a)^2 - 1 = 0
a^2 + 4a^2 + 4a^2 - 1 = 0
9a^2 = 1
a^2 = 1/9
a = 1/3 or -1/3

Substitute -2a for the value of b in the equation (1)
i.e. a^2 + 2a . -2a + (2a)^2 - 1 = 0
a^2 - 4a^2 + 4a^2 - 1 = 0
a^2 - 1 =0
a = 1 or a=-1

When b =2a,
b = 2 . 1/3 = 2/3
Or b=2 . -1/3 = -2/3

When b = -2a,
b = -2 . 1 = -2

Or b = -2a = 2 . -1
= 2

However, a & b are integers so a = -1 and b = 2 or a =1 or b=-2

 

[vds700]

Since it is devisible by (x-1) and (x-4), substituting x=1 or x=4 will yield P(x)=0
i.e. P(1) = (a . 1 + b)^2 - 1 = 0 ...(1)
And P(4) = (a . 4 +b)^2 -4 = 0...(2)

From the equation (1), a^2 + 2ab + b^2 - 1 = 0
(1) x 4: 4a ^2 + 8ab + 4b^2 - 4 = 0...(3)
From the equation (2), 16a^2 + 8ab + b^@ - 4 = 0
(2) - (3):12a^2 - 3b^2 = 0
4a^2 = b^2
b = 2a
Substitute 2a for the value of b in the equation (1)
i.e. a^2 + 2a . 2a + (2a)^2 - 1 = 0
a^2 + 4a^2 + 4a^2 - 1 = 0
9a^2 = 1
a^2 = 1/9
a = 1/3 or -1/3

b = 2a = 2 . 1/3
= 2/3

Or b = 2a = 2 . -1/3
= -2/3

well done Namu, tacking 4 unit questions in year 11

 

[lyounamu]

well done Namu, tacking 4 unit questions in year 11

You are joking, Andrew. You could do this with a single finger and virtually with an absence of mind & brain.

By the way, I didn't know it was a 4 Unit question. I am assuming that this is harder 3 Unit question...(is it not?)

 

[midifile]

You are joking, Andrew. You could do this with a single finger and virtually with an absence of mind & brain.

By the way, I didn't know it was a 4 Unit question. I am assuming that this is harder 3 Unit question...(is it not?)

Nah. There is a whole 4 unit topic on polynomials, which is pretty much the 3 unit topic plus like two new things and a few proofs.

 

[lyounamu]

Nah. There is a whole 4 unit topic on polynomials, which is pretty much the 3 unit topic plus like two new things and a few proofs.

Ok. thanks

 

[Iruka]

Namu, you lost one of the solutions in the middle of your working out!

If 4a^2 = b^2, then b = 2a or b=-2a.

Btw, the question specifies that a and b are integers (which is why you need to look at the other solution.)

 

[namburger]

Since it is devisible by (x-1) and (x-4), substituting x=1 or x=4 will yield P(x)=0
i.e. P(1) = (a . 1 + b)^2 - 1 = 0 ...(1)
And P(4) = (a . 4 +b)^2 -4 = 0...(2)

From the equation (1), a^2 + 2ab + b^2 - 1 = 0
(1) x 4: 4a ^2 + 8ab + 4b^2 - 4 = 0...(3)
From the equation (2), 16a^2 + 8ab + b^@ - 4 = 0
(2) - (3):12a^2 - 3b^2 = 0
4a^2 = b^2
b = 2a
Substitute 2a for the value of b in the equation (1)
i.e. a^2 + 2a . 2a + (2a)^2 - 1 = 0
a^2 + 4a^2 + 4a^2 - 1 = 0
9a^2 = 1
a^2 = 1/9
a = 1/3 or -1/3

b = 2a = 2 . 1/3
= 2/3

Or b = 2a = 2 . -1/3
= -2/3

Thanks for that.
BTW is your name Nam, coz mine is xD

 

[lyounamu]

Namu, you lost one of the solutions in the middle of your working out!

If 4a^2 = b^2, then b = 2a or b=-2a.

Btw, the question specifies that a and b are integers (which is why you need to look at the other solution.)

Ah yeah.

Hahahaha... should have written on the paper before writing on a laptop. How stupid. Will fix mine soon then.

 

[lyounamu]

Thanks for that.
BTW is your name Nam, coz mine is xD

Mine is Namu.

 

[shaon0]

same to shaon0 <!-- Added by James for Benefactors: 11.30pm, 8 Feb 2005 -->

lol....i got the answer but didn't publish it since i was doing Chemistry.
Thanks

 

[lolokay]

Since it is devisible by (x-1) and (x-4), substituting x=1 or x=4 will yield P(x)=0
i.e. P(1) = (a . 1 + b)^2 - 1 = 0 ...(1)
And P(4) = (a . 4 +b)^2 -4 = 0...(2)

From the equation (1), a^2 + 2ab + b^2 - 1 = 0
(1) x 4: 4a ^2 + 8ab + 4b^2 - 4 = 0...(3)
From the equation (2), 16a^2 + 8ab + b^@ - 4 = 0
(2) - (3):12a^2 - 3b^2 = 0
4a^2 = b^2
b = 2a or b = -2a
Substitute 2a for the value of b in the equation (1)
i.e. a^2 + 2a . 2a + (2a)^2 - 1 = 0
a^2 + 4a^2 + 4a^2 - 1 = 0
9a^2 = 1
a^2 = 1/9
a = 1/3 or -1/3

Substitute -2a for the value of b in the equation (1)
i.e. a^2 + 2a . -2a + (2a)^2 - 1 = 0
a^2 - 4a^2 + 4a^2 - 1 = 0
a^2 - 1 =0
a = 1 or a=-1

When b =2a,
b = 2 . 1/3 = 2/3
Or b=2 . -1/3 = -2/3

When b = -2a,
b = -2 . 1 = -2

Or b = -2a = 2 . -1
= 2

However, a & b are integers so a = -1 and b = 2 or a =1 or b=-2

I'd still say it's quicker to just expand both sides and equate coefficients

 

[shaon0]

I'd still say it's quicker to just expand both sides and equate coefficients

Yea equating out the coefficients is a little quicker....thats how i didn't do it.

 

[lolokay]

actually come to think of it shaon0's method was best anyway

 

[lolokay]

i dont see how you can equate coefficients as its not equal to (x - 1)(x - 4) it's just divisible by both (x - 1) and (x - 4).

edit: you can still do it by multiplying the (x-1)(x-4) by a^2, it just takes more time than solving simultaneously straight away

 

[Volt]

i dont see how you can equate coefficients as its not equal to (x - 1)(x - 4) it's just divisible by both (x - 1) and (x - 4).

heres my solution, a little less messy:

P(x) = (ax + b)^2 - x

P(1) = 0 = (a + b)^2 - 1
0 = (a + b - 1) (a + b + 1)

a + b = 1 or a + b = -1

P(4) = 0 = (4a + b)^2 - 4
0 = (4a + b - 2) (4a + b + 2)

4a + b = 2 or 4a + b = -2

3a - 1 = 2
a = 1, b = -2

3a + 1 = -2
a = -1, b = 2

Not only is it equal to (x - 1)(x - 4) for some x, it is in fact identical to it. This is in fact quite an elementary result. It's what being [edit: exactly divisible] divisible by (x - 1) and (x - 4) means!

Knowing this the problem becomes very trivial.

Expand (x - 1)(x - 4) quickly by using the relationship between roots and coefficients, then rearrange to our required form:

x^2 - 5x + 4
≡ (x^2 - 4x + 4) - x
≡ (x^2 - 2*2x + 2^2) - x [to spell it out]
≡ (x - 2)^2 - x

Now it is obvious that (a, b) = (1, -2) or (-1, 2) by observing that p^2 ≡ (-p)^2.