Question

Find the value of the constant a for which the polynomial x^3 + ax^2 -1 will have -1 as a root. (A root is a value of x such that the polynomial is equal to zero.)

Answer 1

Answer:

${ \bf{f(x) = {x}^{3} + {ax}^{2} - 1 }} \\ { \tt{f( - 1) : {( - 1)}^{3} + a {( - 1)}^{2} - 1 = 0}} \\ { \tt{f( - 1) : a - 2 = 0}} \\ a = 2$

Answer 2

The polynomial function $x^3 + ax^2 -1$ will have -1 as a root at the value of

a = 2.

What is a polynomial function?

A polynomial function exists as a function that applies only non-negative integer powers or only positive integer exponents of a variable in an equation like the quadratic equation, cubic equation, etc.

Given: A root exists at a value of x such that the polynomial exists equivalent to zero.

Let, the polynomial equation be $x^3 + ax^2 -1$

then $\mathbf{f}(\mathbf{x})=\mathbf{x}^{3}+a \mathbf{x}^{2}-\mathbf{1}$

Put, x = -1, then we get

$\mathbf{f}(-1)=(-1)^{3}+\mathrm{a}(-1)^{2}-1=0$

f(-1) = a - 2 = 0

a = 2

Therefore, the value of a = 2.

To learn more about polynomial function

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