It’s actually easy enough to do these but I guess you may be faced with having to explain what you’re doing and perhaps even to understand it so I’ll try to help you with all three.

a) and b) are solved in the same way. We know the coefficients of the polynomial are real which means the complex roots (if there are any) must occur in pairs with equal real and opposite imaginary parts. Why is this?

Well, let’s say the roots (zero points) of a polynomial are at x= a, b, c, … then if we write down the expression (x-a)(x-b)(x-c) … and multiply it all out, it gets us right back to the polynomial. In other words a polymonial expression is completely and uniquely defined by its roots. (sorry to seem patronising if you appreciate this already but, if you don’t, it probably seems a bit wierd – anyway it’s a big part of the usefulness of polymonials in science and engineering and it’s the reason why we’re so fascinated by these roots).

So now we come to the reason for the form of complex roots. When we muliply out all the brackets in the polynomial the imaginary parts must cancel out otherwise we’d get complex coefficients. In order to do this they must occur in pairs like a+ib and a-ib (try multiplying it out yourself – you will see that the imaginary parts in the coefficients disappear) so if you know that one of your roots is 6-5i then there must be another root equal to 6+5i and likewise 18+7i must have a compliment in 18-7i. That’s it.

To find the roots of a cubic equation I believe there is a formula but for higher polynomials than that I never knew of one so I’d often use the following simple method:

Make a sketch of the y=polynomial curve by filling in a few values of x and plotting them against the resulting y values on a sheet of paper. This gives you a fair idea where the roots are (values of x where the curve crosses the x axis). In this case there are roots around 1.9, -0.6 and -1.2 very approximately. Using a calculator or better a small programme in Excel to evaluate the polynomial, you can iterate (fool around) with x values until you hit zero as accurately as you want. That’s the root.

I hope this helped,

Bramble

ok, two times the sum of two numbers is 4. we’ve 2 unknowns, we could call them “x” and “y” two times their sum = 4 so we are in a position to write 2(x+y) = 4 so all of us be attentive to that x+y = 2 *** now all of us be attentive to that the sum of their squares is fifty two so, x^2 + y^2 = fifty two @@@ properly, from the *** equation, we are in a position to organize and discover that y = 2 – x so we could sub that equation into our @@@ equation we are going to get x^2 + (2-x)^2 = fifty two now enhance x^2 + 4 – 4x + x^2 = fifty two assemble like words 2x^2 – 4x + 4 = fifty two deliver each little thing to one ingredient, and use the quadratic formula 2x^2 – 4x – 40 8 = 0 (4 +- sqrt(4 hundred) )/4 = a million +- 5 = 6, and -4 now, y = 2 – x so y ought to equivalent -4, or 6 so which you have 2 solutions for this: x = 6, y = -4, and x = -4, y = 6

i think its like 6+5i, and 18-5i, (im not sure though) for the last

use the quadratic equation but make it equal to x^2

so X^2=(-b+-squrt b^2-4ac)/2a

why are you in algebra if u dont know that. lol. also you should know by now since your in college that you have to write a.) as 6+(-5i)duh