polynomial question (1 Viewer)

[Jkitty]

if x^3+3x=ax(x-1)^2+bx(x-1)+cx+d for all values of x, find the values of a,b,c and d

I don't have the answer

 

[Trebla]

if x^3+3x=ax(x-1)^2+bx(x-1)+cx+d for all values of x, find the values of a,b,c and d

I don't have the answer

You can equate the coefficients or substitute specific values of x (or do both) since the equality is meant to hold for all values of x (and hence it should equal no matter what value of x you substitute in).

When you expand the RHS you get x³ + 3x = ax³ + (2a+b)x² + .....

The coefficient of x³ on the RHS is 'a' and the coefficient of x³ on the LHS is 1. Hence a = 1.

I will now do the substitution approach.

When you substitute x = 0, you get d = 0.
When you substitute x = 1, you get c + d = 4 but d = 0 hence c = 4
When you substitute x = 2, you get 2a + 2b + 2c + d = 14 but a = 1, c = 4, d = 0 hence b = 2

 

[Jkitty]

You can equate the coefficients or substitute specific values of x (or do both) since the equality is meant to hold for all values of x (and hence it should equal no matter what value of x you substitute in).

When you expand the RHS you get x³ + 3x = ax³ + (2a+b)x² + .....

The coefficient of x³ on the RHS is 'a' and the coefficient of x³ on the LHS is 1. Hence a = 1.

I will now do the substitution approach.

When you substitute x = 0, you get d = 0.
When you substitute x = 1, you get c + d = 4 but d = 0 hence c = 4
When you substitute x = 2, you get 2a + 2b + 2c + d = 14 but a = 1, c = 4, d = 0 hence b = 2

thank you soooooooooooo much