Find numbers a, b, and c so that a(x2+2x-3) + b(2x2+2x+3) + c(3x2-2x+1) = 2x2-10x+5
Then (a+2b+3c)x2 + (2a+2b-2c)x + (-3a+3b+c) = 2x2 -10x + 5
Equating coefficients of like powers, we have the system of equations:
a + 2b + 3c = 2
2a + 2b - 2c = -10
-3a + 3b + c = 5
Add -2 times the first equation to the second: -2b - 8c = -14
Add 3 times the first equation to the third: 9b +10c = 11
Divide -2b-8c = -14 by -2: b + 4c = 7
9b + 10c = 11
Add -9 times the first equation to the second: -26c = -52
So, c = 2
Backsubstitute to obtain b = -1 and a = -2.
So, -2p1(x) - 1p2(x) + 2p3(x) = q(x)