Verified answer

well, Terry leave me one hour, as i have a meeting right now

and i will provide you with a full answer ! (DONE 1 hr later !)

p = 5t2 −4t +1

we have

p1 = 2t2 +9t −1,

p2 = 4t +2,

p3 = t2 +3t +6

let's start with the necessary condition :

if (because it' s not sure ...)

p can be written as a linear comb of p1, p2 , p3

then there exists a , b and c

so that

p1 = 2t2 +9t −1,

p2 = 4t +2,

p3 = t2 +3t +6 and p = ap1 + bp2 + cp3

then write the expression by grouping the powers and identify each coefficient this gives :

(2a+c)t2 + (9a+4b+3c)t + (-a+2b+6c) = 5t2 -4t + 1

this leads to 3 equations with three unknowns :

2a +c = 5

9a+4b+3c = -4

-a+2b+6c = 1

replacing c = 5-2a in the 2nd and third gives :

9a+4b+3(5-2a) = -4 ------> 3a + 4b = -19

-a+2b+6(5-2a) = 1 -------> -13a+2b = -29

the resolution (is guaranteed !!)

gives :

a = 39/29

b = - 167/29

c = 67/29

if the solution is unique then we showed that it is possible.

hope it' ll help !!

i' ll be right back, if needed !

----> well, i came back as the final solution was missing !

If v can be written as a linear combination of p1, p2, and p3,then there must be 3 constants, a1, a2, a3 such that

v = a1 * p1 + a2 * p2 + a3 * p3 = 5t2 -4t + 1

So we need to find the constants:

Substitute for p1, p2, p3, multiply out and simplify and then equate coefficients on both sides of the equation. You will have 3 equations in 3 unknowns.

v = 2 * a1 * t2 + 9 * a1 * t - a1 + 4 * a2 *t + 2 *a2 + a3 * t2 + 3*a3*t + 6 a3

v = (2a1 + a3 ) * t2 + (9a1 + 4a2 + 3a3) * t + (6a3 + 2a2 -a1) = 5t2 -4t + 1

equating coefficient of the same powers of t on both sides

2a1 + a3 = 5

9a1 + 4a2 + 3a3 = -4

-a1 + 2a3 + 6a3 = 1

which can be written in matrix form

|2 0 1 | |a1| = 5

|9 4 3 | |a2| = -4

|-1 2 6| |a3| = 1

You can use gaussian eiim to solve.

or from the first equation

2a1 + a3 = 5

a3 = 5 - 2a1 so you can plug this into the other 2 equations to get 2 equations in 2 unknowns and solve for a1 and a2.

v=5t^2-4t+1

5t^2-4t+1=a*p1+bp2+c*p3

5t^2-4t+1=a(2t^2+9t-1)+b(4t+2)+c(t^2+3t+6)

5t^2 -4t+1=t^2(2a+c)+t(9a+4b+3c)+(-a+2b+6c)

comparing equal degree term

5=2a+c

-4=9a+4b+3c

1=-a+2b+6c

solve for a,b,c and get answer