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What non-zero rational number must be placed in the square so that the simplified product of these two binomials is a binomial: $(7t - 10)(3t + \square)$? Express your answer as a mixed number.

 Jul 3, 2021
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i feel like i have answered this before on a different website LOL. anyway.

 

we can set this up as an equation, by denoting the box as any variable -- i will let it be $n$

 

so we have $\left(7t-10\right)\left(5t+n\right) \overset{\left(t-m\right)\left(t+n\right)=t^2+nt-mt-mn }{==========\Rightarrow } 35t^2+7nt-50t-10n $

where 

 

$ y=\underbrace{35t^2}_\text{quadratic term}+ \underbrace{7nt-50t}_\text{linear term}  - \underbrace{10n}_\text{constant term} $

 

 

we can see that adding $n$ makes this a trinomial -- to turn it into a binomial we have to either work the constant term or the linear term by equalising them by 0.

 

 

1) constant term:  $ 10n=0 \implies n=0 $  ;  in our question it was stated that $n\neq 0 \ \  \wedge \ \  n\in \mathbb{Q}   $ so that does not statisfy our conditions

 

2) linear term:  $ 7nt-50t= 0 \implies  7nt=50t \implies n=\frac{50}{7} \overset{\text{as a mixed number}}{=======\Rightarrow} 7\frac{1}{7}  $ 

 

 

that does satisfy our conditions, so that is the value of $n$.

$ n=\boxed{7\frac{1}{7}}$


 

 Jul 3, 2021
edited by Guest  Jul 3, 2021
edited by UsernameTooShort  Jul 3, 2021

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