Expand (1+x)^3. Express (1+i)^3 in the form a+bi. Hence, or otherwise, verify that x = 1+i satisfies the equation: x^3+2*x-4i = 0.

First, we factor out one (x+1). (1+x)^3 = (x+1)^2(x+1)= Then, we expand using the formula (a+b)^2 = a^2 + 2ab +b^2: =(x^2+2x+1)(x+1)= Then we multiply: = x^3 + 2x^2 + x + x^2 + 2x + 1 We sum the terms with the same power and we get: (1+x)^3 = x^3 + 3x^2 + 3x +1. For the next part, we expand using the formula we just calculated: (1+i)^3 = i^3 + 3i^2 + 3i + 1 = We replace i^2 with -1: = -i -3 +3i +1 Next, we bring this to the a+bi form: (1+i)^3 = 2i-2. To check if x = i+1 is a root of that equation, we replace x with i+1 and see if the result is 0: (1+i)^3 + 2(1+i) - 4i = We use the form of (1+i)^3 which we calculated above and expand the bracket: 2i-2 + 2i+2 -4i = 0.

Related Further Mathematics A Level answers

All answers ▸

Integrate x^2sin(x) between -pi and pi


Find the four complex roots of the equation z^4 = 8(3^0.5+i) in the form z = re^(i*theta)


Unfortunately this box is to small to contain the question so please see the first paragraph of the answer box for the question.


How can you find the two other roots of a cubic polynomial if you're given one of the roots (which is a complex number)?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences