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For what values of t can 10x^2+tx+8 be written as the product of two binomials
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We can write the quadratic expression 10x^2 + tx + 8 as:

10x^2 + tx + 8 = 10x^2 + (a+b)x + ab

where a and b are constants that we want to determine, and (a+b)x is the middle term in the quadratic expression.

We can factor 10 as 25 and 8 as 22*2, so we have:

10x^2 + tx + 8 = (2x + c)(5x + d)

where c and d are the constants that we need to determine.

Expanding the right-hand side of this equation, we get:

(2x + c)(5x + d) = 10x^2 + (2d+5c)x + cd

Comparing this to the original expression, we see that:

2d + 5c = t

cd = 8

We can use these equations to solve for c and d in terms of t:

c = (t - 2d)/5

d = 8/c

Substituting d in terms of c in the first equation, we get:

2(8/c) + 5c = t

Multiplying through by c, we get a quadratic equation in c:

16 + 5c^2 = tc

We want this equation to have real solutions for c, so the discriminant must be non-negative:

25t^2 - 80 >= 0

Solving this inequality for t, we get:

t <= -8/5 or t >= 8/5

Therefore, the quadratic expression 10x^2 + tx + 8 can be written as the product of two binomials for all values of t less than or equal to -8/5 or greater than or equal to 8/5.

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