**Task:**

Check whether the polynomials p1(x)=x2+4x, p2(x)=1+x3+x2, p3(x)=x3+x, p4(x)=x2-5 form a basis in the space P3.

**Solution:**

The standard basis in the space P3 look like {1, x, x2, x3}, and they have 4 basis elements. To prove that polynomials p1, p2, p3, p4 form a basis in the space P3 it is sufficient to show that they are linearly independent.

Let’s form a linear combination:

*C***1*** p***1** **(*** x***)*** + C***2*** p***2** **(*** x***)*** + C***3*** p***3** **(*** x***)*** + C***4*** p***4** **(*** x***)*** = C***1*** (x ***2*** +***4*** x) + C***2*** (***1*** + x ***3*** +x ***2*** ) + C***3*** (x ***3*** + x) + C***4*** ( x ***2*** — ***5) =*** C***1*** x ***2*** + C***1** **4*** x + C***2*** + C***2*** x ***3*** + C***2*** x ***2*** + C***3*** x ***3*** + C***3*** x + C***4*** x ***2*** – C***4** **5 = **

*x ***3*** (C***2*** + C***3*** ) + x ***2*** (C***1*** + C***2*** + C***4*** ) + x (***4***C***1*** + C***3 **** *** +*** ***C***2*** – ***5***C***4 )**

This linear combination is equal to zero only if all coefficients are zero at powers of x, so we have come to the system:

The system’s only solution is C1 = C2=C3 = C4 = 0 which means that the polynomials p1, p2, p3, p4 are linearly independent and therefore form a basis in the space P3.

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