Help with math question !?

x³ - 10x² + 21x - 30 = (x - 2)(x² + bx + 15)

so, -10 = b - 2

i.e. b = -8

=> x³ - 10x² + 21x - 30 = (x - 2)(x² - 8x + 15)

so, x³ - 10x² + 21x - 30 = (x - 2)(x - 3)(x - 5)

Note, (x - 2)(x - 3)(x - 5) = x³ - 10x² + 31x - 30....so your 21x should be 31x

Then, length is x - 2, width is x - 3 and height x - 5

:)>

There are a few ways to do this:

You can either do some cubic polynomial division, (explained here: https://www.dlsweb.rmit.edu.au/lsu/content/3_maths... in order to get a quadratic, which can then be factorised to get two more values.

Also, you can get the last number (-30) and find 3 numbers which multiply to give that number. (e.g. 2, -3 and 5, but there are lots of other combinations). We are given that one of the numbers is '2', due to the fact that X-2=0. Therefore, the other two numbers must be {15 and 1} or {5 and 3}, or the negatives of each. You need to put each of these new numbers into X and if the equation results in V(0), the value is correct. If the value for 5 were correct, another length would be (x-5) etc.

Hope my poor explanation helps somewhat