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+B

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.

in progress 0
Charlie 2 days 2021-09-09T09:28:21+00:00 2 Answers 0

Answers ( )

    0
    2021-09-09T09:29:22+00:00

    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).

    0
    2021-09-09T09:29:57+00:00

    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.

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )