Polynomials Help (1 Viewer)

[cutemouse]

Could someone please help me with these questions? They are from Fitzpatrick EX 36(b) Q17, 18 respectively:

17. Find constants a, b, c such that the polynomial p(x) = x³-6x²+11x-13 is expressible as X³+aX+b where X=x-c. Hence show that the requation p(x)=0 has only one real root.

18. Factorise the polynomial P(x)=x⁴+2x³-x²-2x. Find a polynomial Q(x) and the constants a and b such that P(x)=(x²+1).Q(x)+ax+b

 

[azureus88]

(x-c)^3 + a(x-c) + b = x^3 - 6x^2 + 11x - 13
By expanding, x^3 - 3cx^2 + (3c^2 + a)x - (c^3 +ac - b) = x^3 - 6x^2 + 11x - 13

Equating coefficients, 3c=6 so c=2
3(2^2) + a = 11 so a = -1
2^3 +(-1)(2) - b = 13 so b = -7

x^3 - 6x^2 + 11x - 13 = X^3 -X -7 = X(X-1)(X+1) - 7

By finding the turning points of Y=X(X-1)(X+1) and then shifting the curve down by 7, you will find that it only has one root.

18. cant you just divide P(x) by (x^2+1) to find Q(x) and ax+b?

 

[Trebla]

18. Factorise the polynomial P(x)=x⁴+2x³-x²-2x. Find a polynomial Q(x) and the constants a and b such that P(x)=(x²+1).Q(x)+ax+b

P(x) = x⁴+ 2x³ - x² - 2x
= x²(x² + 2x) - (x² + 2x)
= (x² + 2x)(x² - 1)
= x(x + 2)(x + 1)(x - 1)
As for the second part, as mentioned already, simply divide P(x) by (x² + 1) and this finds the quotient and remainder.

 

[cutemouse]

Hmm okay, how would I express this algebraically though? Like what I'd write in an exam? I'm not really a fan of doing graphs to show things as I could make mistakes etc..

By finding the turning points of Y=X(X-1)(X+1) and then shifting the curve down by 7, you will find that it only has one root.

 

[Timothy.Siu]

Hmm okay, how would I express this algebraically though? Like what I'd write in an exam? I'm not really a fan of doing graphs to show things as I could make mistakes etc..

u can make mistakes doing anything...graphs are the easiest way to show things, there might not be another way, it says "hence show that", unless u can see another way to do it

 

[cutemouse]

Everybody has their preferences to what they do; I'm just not a fan of graphs, unless they are required (ie. asked in the question). So, if there's another way it'd be good

 

[Timothy.Siu]

they're so easy, u'll be wasting ur time if u dont for some questions, and u need to know how to graph...its a whole topic, and used it lots of other topics

 

[Trebla]

If you want to show algebraically, you need to find the turning points and show that their y values have the SAME SIGN.

If the y values of the turning points are both positive (or both negative) then there is only one real root.

If the y values of the turning points have opposite sign (one is positive, one is negative) then there are three real roots.

You can visualise this by a quick sketch of the graph. Note that this only applies to cubic polynomials.