How do you factor x^4+2x^3y−3x^2y^2−4xy^3−y^4x4+2x3y−3x2y2−4xy3−y4 without quadratic equation?
2 Answers
Explanation:
Given
now making the substitution
but
=-(y+(1/2-sqrt(5)/2)x)(y+(1/2+sqrt(5)/2)x)(y+(3/2-sqrt(5)/2)x)(y+(3/2+sqrt(5)/2)x)=−(y+(12−√52)x)(y+(12+√52)x)(y+(32−√52)x)(y+(32+√52)x)
Explanation:
Given:
x^4+2x^3y-3x^2y^2-4xy^3-y^4x4+2x3y−3x2y2−4xy3−y4
Since this is a homogeneous polynomial, factoring it is similar to factoring a corresponding quartic polynomial in one variable.
To keep it reasonably tidy, reverse the order of the coefficients and their signs to get:
z^4+4z^3+3z^2-2z-1z4+4z3+3z2−2z−1
= (z^4+4z^3+6z^2+4z+1)-3(z^2+2z+1)+1=(z4+4z3+6z2+4z+1)−3(z2+2z+1)+1
= (z+1)^4-3(z+1)^2+1=(z+1)4−3(z+1)2+1
= t^4-3t^2+1" "=t4−3t2+1 wheret = z+1t=z+1
= t^4-2t^2+1-t^2=t4−2t2+1−t2
= (t^2-1)^2-t^2=(t2−1)2−t2
= ((t^2-1)-t)((t^2-1)+t)=((t2−1)−t)((t2−1)+t)
= (t^2-t-1)(t^2+t-1)=(t2−t−1)(t2+t−1)
= ((z+1)^2-(z+1)-1)((z+1)^2+(z+1)-1)=((z+1)2−(z+1)−1)((z+1)2+(z+1)−1)
= (z^2+2z+1-z-1-1)(z^2+2z+1+z+1-1)=(z2+2z+1−z−1−1)(z2+2z+1+z+1−1)
= (z^2+z-1)(z^2+3z+1)=(z2+z−1)(z2+3z+1)
= ((z+1/2)^2-5/4)((z+3/2)^2-5/4)=((z+12)2−54)((z+32)2−54)
= ((z+1/2)-sqrt(5)/2)((z+1/2)+sqrt(5)/2)((z+3/2)-sqrt(5)/2)((z+3/2)+sqrt(5)/2)=((z+12)−√52)((z+12)+√52)((z+32)−√52)((z+32)+√52)
= (z+1/2-sqrt(5)/2)(z+1/2+sqrt(5)/2)(z+3/2-sqrt(5)/2)(z+3/2+sqrt(5)/2)=(z+12−√52)(z+12+√52)(z+32−√52)(z+32+√52)
Hence:
x^4+2x^3y-3x^2y^2-4xy^3-y^4x4+2x3y−3x2y2−4xy3−y4
=-(y^4+4x^3y+3x^2y^2-2x^3y-x^4)=−(y4+4x3y+3x2y2−2x3y−x4)
=-(y+(1/2-sqrt(5)/2)x)(y+(1/2+sqrt(5)/2)x)(y+(3/2-sqrt(5)/2)x)(y+(3/2+sqrt(5)/2)x)=−(y+(12−√52)x)(y+(12+√52)x)(y+(32−√52)x)(y+(32+√52)x)
