Find t if the expansion of the product of x^3-4x^2+2x-5 and x^2+tx-7 has no x^2 term.
(-4x^2)(-7) + (2x)(tx) + (-5)(x^2)
= x^2(28+2t-5)
so, you need 28+2t-5 = 0
To find the value of t if the expansion of the given expression has no x^2 term, we can expand the product of the two polynomials and then compare the coefficient of the x^2 term to zero.
Let's start by multiplying the two polynomials:
(x^3 - 4x^2 + 2x - 5) * (x^2 + tx - 7)
When we expand this expression using the distributive property, we get:
x^5 + (t - 4)x^4 + (2t - 7)x^3 + (-7t - 28)x^2 + (2tx - 5x - 14x - 35)
To have no x^2 term, the coefficient of x^2 must be zero. From the expanded expression, we can see that the coefficient of x^2 is (-7t - 28).
Therefore, we need to solve the equation:
(-7t - 28) = 0
To find the value of t, we can solve this equation:
-7t = 28
Dividing both sides of the equation by -7, we get:
t = -4
Therefore, the value of t for which the expansion of the product of the two given polynomials has no x^2 term is t = -4.