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Given that 2-3i is a root to the equation z^3+pz^2+qz-13p=0, show that p=-2 and q=5.

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
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Answered by William N. • Maths tutor
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Show (2-3i)^3 can be expressed in the form a+bi where a and b are negative integers.

(2-3i) x (2-3i) = -5-12i. -5-12i x (2-3i) = -46-9i. a=-46, b=-9
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Answered by William N. • Maths tutor
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I'm struggling with approaching questions in Maths, I just don't know where to start. What should I do?

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're a...
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Answered by Gus R. • Maths tutor
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State the nth term of the following sequence: 3, 7, 11, 15, 19

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 the number for n, w...
AL
Answered by Alice L. • Maths tutor
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f(x) = 6x³ + 13x² - 4. Use the factor theorem to show that (x+2) is a factor of f(x)

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
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Answered by Benjamin S. • Maths tutor
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