Question

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)(5t+Box )? Express your answer as a mixed number.

Answer 1

The product would expand to

$(7t-10)(5t+x)=35t^2+7tx-50t-10x=35t^2+(7x-50)t-10x$

This is a trinomial, and the only way to make it a binomial is to cancel out a coefficient using our variable $x$.

So, we can cancel either the linear term or the constant term.

In the first case, we require

$7x-50=0 \iff 7x=50 \iff x=\dfrac{50}{7}=\dfrac{49}{7}+\dfrac{1}{7}=7\dfrac{1}{7}$

In the second case, we require

$-10x=0\iff 10x=0 \iff x=0$

But $x$ must be a non-zero rational number, so this solution is not feasible.

Answer 2

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).