How do I know to put (ax - b) as the linear factor?
Are all factors linear?
How am I supposed to work out that b = 5?
You can say that a polynomial of order n is the product of n linear factors. This is the fundamental theorem of algebra.
For example, a quadratic (of order 2) can be split into 2 linear factors, and a cubic can be split into 3. So a polynomial can always be written as a product of linear factors. However, this doesn’t mean that it only has linear factors. With a cubic, for example, you can multiply two of these factors together to get a quadratic factor, as was done in the question. But remember that this quadratic factor is still just the product of two linear factors. Linear factors are like the ‘building blocks’ of higher order polynomials. So yes, and no! - factors can be linear and non-linear but if you want to fully factorize the expression you need to write it as a product of linear factors.
So the above cubic, 3x^3 - 5x^2 + px - 10 = 0, can be written as the product of 3 linear factors: (ax+b), (cx + d) and (ex + f). If this product is equal to 0:
(ax+b)(cx + d)(ex + f) = 0
Then the roots are -b/a, -c/d and -e/f.
We are given one of these roots in the question: √2i. If the equation equals 0 when x = √2i, then (x - √2i) is a factor. According to the complex conjugate root theorem, (x + √2i) must also be a factor. So the cubic can be rewritten as:
(x - √2i)(x + √2i)(ax + b)
You don’t have to put a and b - you could put r, m, α, γ or whatever symbol you like. They just represent an unknown.
The answer shown is a little convoluted - usually you wouldn’t put the equation in a table like that. But the process is hopefully straightforward if we write it differently:
By looking at this equation and equating the coefficients of like x terms, we can see that for it to be true, a must be equal to 3, b must be equal to 5 and p must be equal to 2a. Now we just need to do some simple algebra to find p:
a = 3
p = 2a
p = 2(3)
p = 6
The answer shown has done basically the same thing, they just put the equation in a table instead of expanding it.