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.
+171
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}}$
