# Find a basis of U, the subspace of P3

## Homework Statement

Find a basis of U, the subspace of P3

U = {p(x) in P3 | p(7) = 0, p(5) = 0}

## The Attempt at a Solution

ax3+bx2+cx+d
p(7)=343a+49b+7c+d=0
p(5)=125a+25b+5c+d=0

d=-343a-49b-7c
d=-125a-25b-5c

ax3+bx2+cx+{(d+d)/2} -->{(d+d)/2}=2d/2=d
(-343a-49b-7c-125a-25b-5c)/2=-234a-37b-6c
ax3+b2+cx-234a-37b-6c
a(x3-234)+b(x2-37)+c(x-6)

basis{x3-234,x2-37,x-6}
dim=3
please check if I m correct or not
and is there a easier way to do it?

also, if
U = {p(x) in P3 | p(7) = 0, p(5) = 0,p(3) = 0,p(1) = 0}
p(x) does not exist?
thanks!

Last edited:

Related Precalculus Mathematics Homework Help News on Phys.org
Mark44
Mentor

## Homework Statement

Find a basis of U, the subspace of P3

U = {p(x) in P3 | p(7) = 0, p(5) = 0}

## The Attempt at a Solution

ax3+bx2+cx+d
p(7)=343a+49b+7c+d=0
p(5)=125a+25b+5c+d=0

d=-343a-49b-7c
d=-125a-25b-5c

ax3+bx2+cx+{(d+d)/2} -->{(d+d)/2}=2d/2=d
(-343a-49b-7c-125a-25b-5c)/2=-234a-37b-6c
ax3+b2+cx-234a-37b-6c
a(x3-234)+b(x2-37)+c(x-6)

basis{x3-234,x2-37,x-6}
dim=3
please check if I m correct or not
and is there a easier way to do it?

also, if
U = {p(x) in P3 | p(7) = 0, p(5) = 0,p(3) = 0,p(1) = 0}
p(x) does not exist?
thanks!
No, that's not right.

Your subspace of P3 is the set of 3rd degree polynomials p(x) such that p(7) = 0 and p(5) = 0. That means that any polynomials in this space must have factors of (x -7) and (x - 5). So any polynomial in this subspace must satisfy (x - 7)(x - 5)(ex + f) = 0, with e and f being arbitrary.

There's more to do, but this will get you going in the right direction.

so the basis would be {(x - 7)(x - 5)x,(x - 7)(x - 5)}? dim=2

U = {p(x) in P3 | p(7) = 0, p(5) = 0,p(3) = 0,p(1) = 0}
does p(x) exist?
it has 4 degree(x - 7)(x - 5)(x - 3)(x - 1), but P3 is a set of third degree, so it does not exist
is that right?

Last edited:
Mark44
Mentor
For the first question: yes.
For the second question: there is no 3rd degree polynomial with four zeroes.

can i write does not exist?
thanks!

Mark44
Mentor
Sure. "There is no ..." and "does not exist" mean the same thing.

Mark44
Mentor
For the first question: yes.
For the second question: there is no 3rd degree polynomial with four zeroes.
Correction (and thanks to vela!): There is no 3rd degree polynomial with exactly four zeroes, but p(x) = 0x3 + 0x2 + 0x + 0 satisfies p(7) = p(5) = p(3) = p(1) = 0.

does that mean the basis exist {1,x,x2,x3}?

Mark44
Mentor
1) If U is as described in post #1, what is a basis for U? (Already answered)
2) Is there a polynomial in P3 such that p(7) = p(5) = p(3) = p(1) = 0? Yes, the zero polynomial.

Yes.
But the question in my assignment is
Find a basis of U, the subspace of P3
d) U = {p(x) in P3 | p(7) = 0, p(5) = 0,p(3) = 0,p(1) = 0}
and I wrote no basis

Mark44
Mentor