To perform this check, you must use the following theorem: [tex]a[/tex] is a lower bound for the zeroes of [tex]f(x)[/tex] if, when divide [tex]f(x)[/tex] by [tex](x-a)[/tex], the quotient and the remainter alternate signs.
Since the long division yields the result
[tex]\dfrac{x^4+x^3-11x^2-9x+18}{x+1}=x^3-11x+2+\dfrac{16}{x+1}[/tex]
You can see that the signs don't alternate (the last two terms are positive). So, -1 is not a lower bound.
In fact, the actual roots of the polynomial are -3, -2, 1, 3, so as you can see there are roots smaller than -1.
The value of x = -1 is not the lower bound for the zeroes of the function given.
A function is a law that relates a dependent and an independent variable.
The function is
f(x) = x⁴ +x³ -11 x² -9 x+18
The value x=-1 will be a lower bound only when the function when divided by (x+1) yields alternate sign of quotient and remainder.
(x⁴ +x³ -11 x² -9 x+18)/(x+1) = x³-11x +2 is the quotient and 16 is the remainder, and they do not have alternate signs
so, x= -1 is not the lower bound.
The roots of the polynomial are -3, -2, 1, 3, which is less than -1.
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