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How do you factor t2-18t plus 81? - Answers

Hi. To see if you can factorise this equation, which I'm guessing is meant to be t squared, you should use the quadratic formula which is b2-4ac. in your question ax2+bx+c (x is the same as t) a = 1 as it is the co-efficient of t2 (the coefficient is the number in front of the t2) b = -18 (this is because it is in front of the t) c = 81 so (-18)2- 4*1*81 = 0 as this equals 0 this means that there is only one solution so when your factorise both of the brackets will contain the same thing. so if you say what two numbers multiply together to get 81. say 9*9 consider these numbers for adding together to the b value. if you add these you get 18. this is no the b value. therefore we can try -9 and -9. adding these together gives you -18 which is the b value and it works for the multiply part aswell as the minus cancel. so from here we can say that t2-18t+81 = (t-9)(t-9) or could be written as (t-9)2 hope this helps you.



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How do you factor t2-18t plus 81? - Answers

https://math.answers.com/other-math/How_do_you_factor_t2-18t_plus_81

Hi. To see if you can factorise this equation, which I'm guessing is meant to be t squared, you should use the quadratic formula which is b2-4ac. in your question ax2+bx+c (x is the same as t) a = 1 as it is the co-efficient of t2 (the coefficient is the number in front of the t2) b = -18 (this is because it is in front of the t) c = 81 so (-18)2- 4*1*81 = 0 as this equals 0 this means that there is only one solution so when your factorise both of the brackets will contain the same thing. so if you say what two numbers multiply together to get 81. say 9*9 consider these numbers for adding together to the b value. if you add these you get 18. this is no the b value. therefore we can try -9 and -9. adding these together gives you -18 which is the b value and it works for the multiply part aswell as the minus cancel. so from here we can say that t2-18t+81 = (t-9)(t-9) or could be written as (t-9)2 hope this helps you.



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https://math.answers.com/other-math/How_do_you_factor_t2-18t_plus_81

How do you factor t2-18t plus 81? - Answers

Hi. To see if you can factorise this equation, which I'm guessing is meant to be t squared, you should use the quadratic formula which is b2-4ac. in your question ax2+bx+c (x is the same as t) a = 1 as it is the co-efficient of t2 (the coefficient is the number in front of the t2) b = -18 (this is because it is in front of the t) c = 81 so (-18)2- 4*1*81 = 0 as this equals 0 this means that there is only one solution so when your factorise both of the brackets will contain the same thing. so if you say what two numbers multiply together to get 81. say 9*9 consider these numbers for adding together to the b value. if you add these you get 18. this is no the b value. therefore we can try -9 and -9. adding these together gives you -18 which is the b value and it works for the multiply part aswell as the minus cancel. so from here we can say that t2-18t+81 = (t-9)(t-9) or could be written as (t-9)2 hope this helps you.

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      Hi. To see if you can factorise this equation, which I'm guessing is meant to be t squared, you should use the quadratic formula which is b2-4ac. in your question ax2+bx+c (x is the same as t) a = 1 as it is the co-efficient of t2 (the coefficient is the number in front of the t2) b = -18 (this is because it is in front of the t) c = 81 so (-18)2- 4*1*81 = 0 as this equals 0 this means that there is only one solution so when your factorise both of the brackets will contain the same thing. so if you say what two numbers multiply together to get 81. say 9*9 consider these numbers for adding together to the b value. if you add these you get 18. this is no the b value. therefore we can try -9 and -9. adding these together gives you -18 which is the b value and it works for the multiply part aswell as the minus cancel. so from here we can say that t2-18t+81 = (t-9)(t-9) or could be written as (t-9)2 hope this helps you.
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