22 Jun Answer: ##8(x-1)(x^2+x+1)## To factorize the expression firs
Answer: ##8(x-1)(x^2+x+1)## To factorize the expression first we remove any obvious common factors of all the terms. Here 8 divides every term so we can extract the 8: ##8(x^3-1)## Next we see if there are any factors of the expression ##(x^3-1)##. If there is a solution ##alpha## to the equation ##x^3-1 = 0## then ##(x-alpha)## will divide ##(x^3 – 1)## (This is called the ‘factor theorem’.) Here ##x = 1## is an obvious solution so we can take out ##(x-1)## as a factor ##8(x-1)f(x)## where ##(x-1)f(x) = (x^3 – 1)## and ##f(x)## is a polynomial we must work out. ##f(x)## must be quadratic (degree 2) since ##x^3-1## has degree 3 so we can write it as a general quadratic: ##f(x) = ax^2 + bx + c## where we need to work out the values of a b and c. Multiplying out the brackets: ##x^3 – 1 = (x-1)f(x) = (x-1)(ax^2 + bx + c)## ##= ax^3 + bx^2 + cx – ax^2 – bx – c## ##=ax^3 + (b-a)x^2 + (c-b)x -c## and by comparing the coefficients of the powers of x we see that ##a=1## ##b-a = 0## ##c-b=0## and ##c=1## implying that ##b=1##. So we have our newly factored polynomial: ##8(x-1)(x^2 + x + 1)## Can we factorize this further? In other words are there any solutions to the equation ##x^2 + x + 1 = 0##? By using the quadratic formula we find that the solutions are given by: ##x = frac(-1 + sqrt(-3))(2)## or ##frac (-1 – sqrt(-3))(2)## Both of these solutions however involve taking the root of a negative number which does not give a real number. Therefore the equation has been factorized as much as possible in the real numbers and our job is complete: ##8x^3 – 8 = 8(x-1)(x^2 + x + 1) quadquadquadquadquadquadsquare##
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