Given the polynomials $q\left(x\right)=-{x}^{3}+3{x}^{2}-x+5,r\left(x\right)=-4{x}^{3}+7{x}^{2}-x+10$ and $u\left(x\right)=-5{x}^{3}+8{x}^{2}$.

Now, a polynomial p(x) is in span of {q,r,u} if there exist nonzero constants a,b and c such that $p\left(x\right)=aq+br+cu$

The polynomial is ${p}_{1}\left(x\right)=-3{x}^{2}-3x+10$

Then,

${p}_{1}=aq+br+cu$

$-3{x}^{2}-3x+10=a(-{x}^{3}+3{x}^{2}-x+5)+b(-4{x}^{3}+7{x}^{2}-x+10)+c(-5{x}^{3}+8{x}^{2}+10)-3{x}^{2}-3x+10$

$=(a-4b-5c){x}^{3}+(3a+7b+8c){x}^{2}+(-a-b)x+5a+10b+10c$

Equate the coefficients of both sides and find the values of a, b and c.

$a-4b-5c=0$ (1)

$3a+7b+8c=-3$ (2)

$-a-b=-3$ (3)

$5a+10b+10c=10$ (4)

From equation (3),

$a=-3+b$

$a=3-b$

Substitute $b=3-a$ in equation (1) and equation (2) and (3)

$-(3-b)-4b-5c=0$ (4)

$3(3-b)+7b+8c=-3$ (5)

$5(3-b)+10b+10c=10$ (6)

From equation (4),

$-3+b-4b-5c=0$

$-3-3b-5c=0$

$-3b=5c+3$

$b=-\frac{5c+3}{3}$

Substitute $b=-\frac{5c+3}{3}$ and in equation (5) and (6)

$-9+4(-\frac{5c+3}{3})+8c=-3$ (7)

$15+5(-\frac{5c+3}{3})+10c=10$

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