Respuesta :
Answer:
[tex] p(x) = ax^3 +bx^2 +cx +d [/tex]
Where a,b,c,d represent constants.
For case given we have this:
[tex] p(1)= a + b+c+d [/tex]
And we need that [tex] p(1)=0[/tex]
[tex] p(1)= a + b+c+d=0[/tex]
That's satisfied if [tex] d = -(a+b+c)[/tex]
[tex] p(x) = ax^3 +bx^2 +cx -(a+b+c) [/tex]
Where [tex] a,b,c,d \in R[/tex]
So then a possible basis for this case would be:
[tex] B= [x^3 -1, x^2 -1 , x-1][/tex]
And we can check that the spans for the subspace is linearly independent.
Step-by-step explanation:
We want to find a basis for the subspace of all polynomials with degree [tex] \leq 3[/tex] and we need to satisfy the condition that [tex] p(1) =0[/tex]
Our general expression for an element of the space described above is given by this equation:
[tex] p(x) = ax^3 +bx^2 +cx +d [/tex]
Where a,b,c,d represent constants.
For case given we have this:
[tex] p(1)= a + b+c+d [/tex]
And we need that [tex] p(1)=0[/tex]
[tex] p(1)= a + b+c+d=0[/tex]
That's satisfied if [tex] d = -(a+b+c)[/tex]
So we want a basis for the subsapce of all the polynomials on this form:
[tex] p(x) = ax^3 +bx^2 +cx -(a+b+c) [/tex]
Where [tex] a,b,c,d \in R[/tex]
So then a possible basis for this case would be:
[tex] B= [x^3 -1, x^2 -1 , x-1][/tex]
And we can check that the spans for the subspace is linearly independent.
