Find t if x=-1 is root of f=x^3-4x^2+2tx-5?

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-You should remember that substituting the value of root in polynomial equation, this value verifies the equation such that:

`f(-1) = 0`

Substituting -1 for x in polynomial equation yields:

`f(-1) = (-1)^3 - 4(-1)^2 + 2t*(-1) - 5`

`f(-1) = -1 - 4 - 2t - 5`

Since `f(-1) = 0` , hence `-1 - 4 - 2t - 5 = 0` such that:

`-2t - 10 = 0 =gt -2t = 10 =gt t = 10/(-2) =gt t = -5`

**Hence, evaluating t under given conditions yields `t = -5` .**

**Sources:**

We have the function as

f(x)=x^3-4x^2+2tx-5, since -1 is a root of f(x) then, f(-1) must be 0

that is f(-1)=0, now let's plugh in the value of x in the given function

x^3-4x^2+2tx-5 = 0

(-1)^3-4(-1)^2+2(-1)t-5=0

-1-4*1-2t-5=0

-1-4-2t-5=0

-10-2t=0

-2t=10

t=10/-2 = -5

so the value of t at x=-1 That is f(-1) is -5

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