Respuesta :
The product would expand to
[tex](7t-10)(5t+x)=35t^2+7tx-50t-10x=35t^2+(7x-50)t-10x[/tex]
This is a trinomial, and the only way to make it a binomial is to cancel out a coefficient using our variable [tex]x[/tex].
So, we can cancel either the linear term or the constant term.
In the first case, we require
[tex]7x-50=0 \iff 7x=50 \iff x=\dfrac{50}{7}=\dfrac{49}{7}+\dfrac{1}{7}=7\dfrac{1}{7}[/tex]
In the second case, we require
[tex]-10x=0\iff 10x=0 \iff x=0[/tex]
But [tex]x[/tex] must be a non-zero rational number, so this solution is not feasible.
Answer:
7 1/7.
Step-by-step explanation:
The middle 2 numbers in the expansion will be cancelled out if one of them is + 50 t so the required rational number is 50/7:
(7t - 10)(5t + 50/7) = 35t^2 - 50t + 350/7 t - 500/7)
= 35t^2 - 50t + 50t - 500/7)
= 35t^2 - 500/7)
So 50 / 7 = 7 1/7 (answer).
