Question and Answers Forum

All Questions      Topic List

Matrices and Determinants Questions

Previous in All Question      Next in All Question      

Previous in Matrices and Determinants      Next in Matrices and Determinants      

Question Number 10268 by j.masanja06@gmail.com last updated on 01/Feb/17

Answered by sandy_suhendra last updated on 01/Feb/17

(x−3)(x+5)(x−1)+6+42−6(x+5)+6(x−3)+7(x−1)=0 x^3 +x^2 −10x+8=0 (x−1)(x^2 +2x−8)=0 (x−1)(x−2)(x+4)=0 x=1 or x=2 or x=−4

(x−3)(x+5)(x−1)+6+42−6(x+5)+6(x−3)+7(x−1)=0x3+x2−10x+8=0(x−1)(x2+2x−8)=0(x−1)(x−2)(x+4)=0x=1orx=2orx=−4

Answered by arge last updated on 04/Feb/17

x−3 1 −1 −7 x+5 −1 −6 6 x−1 x−3 1 −1 −7 x+5 −1 (x−3)(x+5)(x−1)+(−7×6×−1)+(−6×1×−1)−{−1(x+5)(−6)+[−1(6)(x−3)+(x−1)(1)(−7)}=0 x^3 +x^2 −24x+22=0 (x−1)(x^2 +2x−22)=0 x=1

x−31−1−7x+5−1−66x−1x−31−1−7x+5−1(x−3)(x+5)(x−1)+(−7×6×−1)+(−6×1×−1)−{−1(x+5)(−6)+[−1(6)(x−3)+(x−1)(1)(−7)}=0x3+x2−24x+22=0(x−1)(x2+2x−22)=0x=1

Terms of Service

Privacy Policy

Contact: info@tinkutara.com