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Substitute 2-3i into equation using part i (2-3i)³=-46-9i.  -46-9i+p(-5-12i)+q(2-3i)-13p=0. -46-18p+2q-9i-12pi-3iq=0. Real: -46-18p+2q=0 and Imaginary: -9-12p-3q=0. p=-2, q=5

(2-3i) x (2-3i) = -5-12i.  -5-12i x (2-3i) = -46-9i.  a=-46, b=-9

My approach will take you step by step through each type of question you find difficult. Maths (quite usefully) is very procedural. This means if you follow the instructions step by step you'll find you'r...

Start by labelling each term, e.g. 3 is the 1st, 7 is the second etc

Find the difference between each term, in this case +4 so we know our nth term will start with 4n 

Now we substitute in t...

If x+2 is a factor implies x=-2 is a solution

Sub in x=-2 : f(-2) = 6*(-2)³ + 13*(-2)² -4 = 6*-8 +13*4 - 4 = 0

f(-2) = 0 therefore x+2 is a solution