Question

Find all combinations of a,b,c and d which makes the following equality an identity
\((x-a)^2(x-b)^2=x^4-2x^3-3x^2+cx+d\) where a 12 years ago

Answer 1

consider:$$(x-a)(x-b)=x^2-(a+b)x+ab\\(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc\\(x-a)(x-b)(x-c)(x-d)=x^4-(a+b+c+d)x^3+\\\ \ \ \ \ \ (ab+ac+ad+bc+bd+cd)x^2-(abc+abd+acd+bcd)x+abcd$$

Answer 2

hence \((x-a)^2(x-b)^2\) gives:$$x^4-2(a+b)x^3+(a^2+4ab+b^2)x^2-(a^2b+a^2b+ab^2+ab^2)x+a^2b^2$$

Answer 3

so by inspection of coefficients we determine:$$a+b=1\\a^2+4ab+b^2=-3$$

Answer 4

note \(a+b=1\) means \((a+b)^2=a^2+2ab+b^2=1\) hence:$$a^2+4ab+b^2=-3\\1+2ab=-3\\2ab=-4\\ab=-2$$

Answer 5

@ujjwal do \(a,b\) have to be integers? if so, then consider factorization...$$-1\times2=-2\\-2\times1=-2$$which gives two solutions for \((a,b)\) namely \((-1,2),(-2,1)\)

Answer 6

if not then there is an infinite number of combinations!

Answer 7

they have to be integers and a<b ..
Thanks!

Answer 8

a+b=1 so only 2,-1

Answer 9

by the way, @ujjwal, I made use of http://en.wikipedia.org/wiki/Vieta's_formulas which relate the roots of polynomials to the coefficients

Answer 10

oops @hartnn good eye.. \((-2,1)\) satisfies \(a^2+4ab+b^2=-3\) but gives \(a+b=-1\) so only \((-2,1)\) works