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How do you solve an equation by factorising? - Answers

Let A and B be any number. Let's say you have (5-A)(10-B) = 0. For this to be true, either (5-A) or (10-B) must be zero for the the answer to equal zero. Essentially what you are saying is 0(10-B)=0 or (10-A)0=0. This let's you solve for A and B. You can just set each "factor" to zero and solve for the variable. 5-A=0 A=5 10-B=0 B=10 Another "un-factored" example: x2 - 10x = -25 x2 - 10x + 25 = 0 <--- the factors have to be equal to zero for this to work You need two numbers that add to negative 10 and multiply to 25. (-5) + (-5) = -10 (-5) x (-5) = 25 Therefore: (x-5)(x-5)=0 And x can only be equal to 5, x=5



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How do you solve an equation by factorising? - Answers

https://math.answers.com/math-and-arithmetic/How_do_you_solve_an_equation_by_factorising

Let A and B be any number. Let's say you have (5-A)(10-B) = 0. For this to be true, either (5-A) or (10-B) must be zero for the the answer to equal zero. Essentially what you are saying is 0(10-B)=0 or (10-A)0=0. This let's you solve for A and B. You can just set each "factor" to zero and solve for the variable. 5-A=0 A=5 10-B=0 B=10 Another "un-factored" example: x2 - 10x = -25 x2 - 10x + 25 = 0 <--- the factors have to be equal to zero for this to work You need two numbers that add to negative 10 and multiply to 25. (-5) + (-5) = -10 (-5) x (-5) = 25 Therefore: (x-5)(x-5)=0 And x can only be equal to 5, x=5



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https://math.answers.com/math-and-arithmetic/How_do_you_solve_an_equation_by_factorising

How do you solve an equation by factorising? - Answers

Let A and B be any number. Let's say you have (5-A)(10-B) = 0. For this to be true, either (5-A) or (10-B) must be zero for the the answer to equal zero. Essentially what you are saying is 0(10-B)=0 or (10-A)0=0. This let's you solve for A and B. You can just set each "factor" to zero and solve for the variable. 5-A=0 A=5 10-B=0 B=10 Another "un-factored" example: x2 - 10x = -25 x2 - 10x + 25 = 0 <--- the factors have to be equal to zero for this to work You need two numbers that add to negative 10 and multiply to 25. (-5) + (-5) = -10 (-5) x (-5) = 25 Therefore: (x-5)(x-5)=0 And x can only be equal to 5, x=5

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      Let A and B be any number. Let's say you have (5-A)(10-B) = 0. For this to be true, either (5-A) or (10-B) must be zero for the the answer to equal zero. Essentially what you are saying is 0(10-B)=0 or (10-A)0=0. This let's you solve for A and B. You can just set each "factor" to zero and solve for the variable. 5-A=0 A=5 10-B=0 B=10 Another "un-factored" example: x2 - 10x = -25 x2 - 10x + 25 = 0 <--- the factors have to be equal to zero for this to work You need two numbers that add to negative 10 and multiply to 25. (-5) + (-5) = -10 (-5) x (-5) = 25 Therefore: (x-5)(x-5)=0 And x can only be equal to 5, x=5

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