need-help-with-step-by-step-thanks-x-x-4-x-1-2x-x-4- Tinku Tara June 3, 2023 Algebra 0 Comments FacebookTweetPin Question Number 7234 by WagST last updated on 17/Aug/16 needhelpwithstepbystep.thanks.x(x+4)(x−1)=2x(x+4) Answered by Yozzia last updated on 17/Aug/16 x(x+4)(x−1)=2x(x+4)(∗)Adding−2x(x+4)onbothsidesof(∗)givesx(x+4)(x−1)−2x(x+4)=2x(x+4)−2x(x+4)∵2x(x+4)−2x(x+4)=0∴x(x+4)(x−1)−2x(x+4)=0Bythedistributivelawa(b+c)=ab+bc⇒x(x+4)[(x−1)−2]=0⇒x(x+4)[x−3]=0Now,ingeneral,aproducta×b×c×…×z=0iffatleastoneofa,b,c,…,ziszero.Henceforx(x+4)(x−3)=0atleastoneofx,(x+4)or(x−3)iszero.∴x=0orx=−4orx=3. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: S-t-1-n-1-t-t-k-k-n-k-gt-1-Solve-for-S-Next Next post: Question-72773 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![x(x+4)(x−1)=2x(x+4) (∗) Adding −2x(x+4) on both sides of (∗) gives x(x+4)(x−1)−2x(x+4)=2x(x+4)−2x(x+4) ∵ 2x(x+4)−2x(x+4)=0 ∴ x(x+4)(x−1)−2x(x+4)=0 By the distributive law a(b+c)=ab+bc ⇒ x(x+4)[(x−1)−2]=0 ⇒ x(x+4)[x−3]=0 Now, in general, a product a×b×c×...×z=0 iff at least one of a,b,c,...,z is zero. Hence for x(x+4)(x−3)=0 at least one of x, (x+4) or (x−3) is zero. ∴ x=0 or x=−4 or x=3.](https://www.tinkutara.com/question/Q7235.png)