Question

Let t represent an unknown number. What is the value

oft such that: t^2 x^3– 3tx + 6 is divisible by x+3.

Answer 1

There are two possible answers:

t = 2/3 or t = -1/3

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Step-by-step explanation:

The expression x+3 is the same as x-(-3). It is of the form x-k

Recall that by the remainder theorem, we know that p(x) is divisible by (x-k) if and only if p(k) = 0.

In this case, k = -3. Plug this into the p(x) function and we get

p(x) = t^2*x^3 - 3t*x + 6

p(-3) = t^2*(-3)^3 - 3t*(-3) + 6

p(-3) = -27t^2 + 9t + 6

Set that equal to 0

p(-3) = 0

-27t^2 + 9t + 6 = 0

-3(9t^2 - 3t - 2) = 0

9t^2 - 3t - 2 = 0

Now apply the quadratic formula to solve for t

`t = (-b\pm√(b^2-4ac))/(2a)\\\\t = (-(-3)\pm√((-3)^2-4(9)(-2)))/(2(9))\\\\t = (3\pm√(81))/(18)\\\\t = (3\pm9)/(18)\\\\t = (3+9)/(18)\ \text{ or } \ t = (3-9)/(18)\\\\t = (12)/(18)\ \text{ or } \ t = (-6)/(18)\\\\t = (2)/(3)\ \text{ or } \ t = -(1)/(3)\\\\`

We get two possible solutions for t.