Question

The polynomial P(x)=x^4 - 4x^3 + hx^2 - 6x + 2 has a factor of (x-m)^2 . Find value of m and h

Answer 1

Ah, I like this problem :-)

You can solve it by remembering that if \((x-m)\) is a factor of \(P(x)\), then \(P(m) = 0\), and the rational root theorem says that \(m\) will be a positive or negative factor of the constant term in \(P(x)\). Try evaluating \(P(x)\) for various factors of 2, and then solve the resulting equation for \(h\). Check the result to see if \(P(x)\) has \((x-m)^2\) as a factor when you plug in the prospective value of \(h\).

Answer 2

we are to assume that \(m\) is an integer?

Answer 3

Ya

Answer 4

then i guess it has to be either 1 or -1
@whpalmer4 right?

Answer 5

Yep!

Answer 6

if \(x=1\) then it must be
\[x^4 - 4x^3 + hx^2 - 6x + 2=(x^2-2x+1)(something)\] and you can figure out what the "something" is

Answer 7

whpalmer can you work out the procedure , I can not really understand !

Answer 8

Hmm, @satellite73 I think you made an error somewhere in your factoring or division...

Answer 9

maybe, but i didn't divide i just wrote what you have to get if you have a factor of \((x-1)^2\)

Answer 10

@satellite73 that doesn't multiply out correctly to give a constant term of 2 in P(x)

Answer 11

1*1 = 1, not 2 :-)

Answer 12

as \((x-1)^2=x^2-2x+1\) i must factor as
\[x^4 - 4x^3 + hx^2 - 6x + 2=(x^2-2x+1)(x^2+bx+2)\]

Answer 13

oh, okay, you're looking at the other half, never mind...

Answer 14

it's early, too much blood in the caffeine :-)

Answer 15

Well, let's try x = 1 in P(x):
\[P(x) = (1)^4-4(1)^3+h(1)^2-6(1)+2 = 1-4+h-6+2 = h-7 = 0\]
\[h = 7\]

Answer 16

\[P(x) = x^4-4x^3+7x^2-6x+2\]

Answer 17

we can find \(b\) by noting that when you multiply you get \(bx-4x=-6x\)=

Answer 18

as usual your method looks faster

Answer 19

but what if \(x=-1\) ?

Answer 20

Then we would have h = -13
\[P(x) = x^4-4x^3-13x^2-6x+2\] which doesn't have \((x+1)^2\) as a factor

Answer 21

just dumb luck that we tried the right solution first :-)

Answer 22

I have a question , m must be 1/-1? Can not be 2 or others?

Answer 23

Well, let's suppose the value is m=2. Then \((x-2)^2\) must be a factor of \(P(x)\). However, if we expand \((x-2)^2\):
\[(x-2)(x-2) = x^2-2x-2x+4 = x^2-4x+4\]What can we multiply that by which has integer coefficients and end up with a constant term of 2?

Answer 24

Remember, the constant term is always the product of all of the constant terms multiplied together:

\[(x-a)(x-b)(x-c)(x-d) = a b c d-a b c x-a b d x-a c d x-b c d x+a b x^2+\]\[a c x^2+b c x^2+a d x^2+b d x^2+c d x^2-a x^3-b x^3-c x^3-d x^3+x^4\]

Answer 25

if \(a = b = 2\), and \(a * b * c * d = 2\), and \(a,b,c,d\) are all integers, we need \(c*d = 1/2\) in order to make that work, but there aren't any integers that fit the bill...

Answer 26

Thank for the explanation!

Answer 27

does it make sense? :-)

Answer 28

Ya

Answer 29

good! As my esteemed colleague rightly asked, we did assume an integer solution here...but there's no harm in assuming that, at worst we don't find one and have to try a different approach. The RRT told us what candidates we had to work with if an integer solution exists.

Answer 30

Just be very glad the constant term was 2, and not say 24, and that the leading coefficient was 1! :-)