2x² + C = A (x - 1)² + B (x - 1) - 9

Since this is going to be an identity,

1. The corresponding coefficients and the constant term must be equal.

2. The identity works for any value of x.

However, we cannot straight away say that C = -9 since there are actually hidden constants within the bracketed terms.

I think that C is -11 , Ive figured that out I just don’t know how to find the other two unknown values

2x² + C = A (x - 1)² + B (x - 1) - 9

Letting x = 1 (why do you think I did so?),

2 (1)² + C = A (1 - 1)² + B (1 - 1) - 9
2 + C = -9
C = -11

2x² - 11 = A (x - 1)² + B (x - 1) - 9
2x² - 11 = A (x - 1) (x - 1) + B (x - 1) - 9

Because the only possible combination of x² on the right-hand side comes from A multiplied by x multiplied by another x, we can compare the terms in x².

2x² = A (x) (x)
2x² = Ax²
2 = A

2x² - 11 = 2 (x - 1)² + B (x - 1) - 9

Now we have one unknown left, so it's free and easy.

Letting x = 0,
2 (0)² - 11 = 2 (0 - 1)² + B (0 - 1) - 9
-11 = 2 - B - 9
-11 = -B - 7
-4 = -B
4 = B

So, A = 2, B = 4 and C = -11.

Alternatively, we can expand everything and compare the corresponding cofficients and constants.

2x² + C = A (x - 1)² + B (x - 1) - 9
2x² + C = A (x² - 2x + 1) + B (x - 1) - 9
2x² + C = Ax² - 2Ax + A + Bx - B - 9
2x² + C = Ax² + (B - 2A)x + A - B - 9

Comparing terms in x²,
2x² = Ax²
2 = A

Comparing terms in x (LHS has no term in x so we say 0x or simply 0),
0x = (B - 2A)x
0 = B - 2A
0 = B - 2(2)
0 = B - 4
4 = B

Comparing constants,
C = A - B - 9
C = 2 - 4 - 9
C = -11

Earlier on I wrote this

1. The corresponding coefficients and the constant term must be equal.

2. The identity works for any value of x.

Letting x = number makes use of the second point which I mentioned - the identity should hold for all values of x. However, some choices for x are better than others.

In the worst case scenario, if you choose poor values of x (eg x = 3, x = 4, x = 5, ...), we will obtain three simultaneous equations in three unknowns to solve for. This will still get you the values of A, B and C, though it's quite inefficient.

Comparing coefficients or using my alternative approach makes use of the first point, because identities are supposed to work for every single value of x. If the coefficients are not the same, then the result will not hold for all values of x (and will usually hold only for maybe one, two or a finite number of solutions).

As you proceed with these questions, you can feel free to practise either approach and adopt the most appropriate or comfortable approach. In most questions, I adopt a mix of each approach in a single question to obtain the quickest solution.

After you find C, if you have difficulty finding the other two values, you can even choose two random values of x, though you will obtain a pair of simultaneous equations in two unknowns A and B. It's tedious to solve the resulting equations, but it's the most obvious approach.

I will illustrate in a moment.

2x² - 11 = A (x - 1)² + B (x - 1) - 9

This is a rough idea to illustrate what happens in your worst case scenario.

You have two unknowns, so you need at least two equations to solve them (except in an extreme minority of questions involving square roots). I will let x = 3 and x = 4 for this.

(You can try with x = 0 and x = 2 later on, with the same values of A and B)

Letting x = 3,
2 (3)² - 11 = A (3 - 1)² + B (3 - 1) - 9
18 - 11 = 4A + 2B - 9
7 = 4A + 2B - 9
16 = 4A + 2B
8 = 2A + B
8 - 2A = B

Letting x = 4,
2 (4)² - 11 = A (4 - 1)² + B (4 - 1) - 9
32 - 11 = 9A + 3B - 9
21 = 9A + 3B - 9
30 = 9A + 3B
10 = 3A + B

So, from there on,

10 = 3A + 8 - 2A
10 = 8 + A
2 = A

and B = 8 - 2(2) = 4

as before.

thank you very much ! you’ve been a tremendous help <3