Question

Suppose that \(f(x)=4 x^{3}-11 x^{2}-26 x+24\). Find the zeros of \(f(x-2)\).

Short answer

Solve \(4y^3 + 13y^2 - 22y - 40 = 0\) for \(y\), then substitute back \( x = y + 2 \).

Step-by-step solution

Understand the problem

The task is to find the zeros of the transformed function. First, recognize that the function given is a cubic polynomial: \(f(x) = 4x^3 - 11x^2 - 26x + 24\). We need to find the zeros of \(f(x-2)\).

Substitute \(x - 2\) into the function

To obtain \(f(x-2)\), substitute \(y = x - 2\) into \(f(y)\). First rewrite the function in terms of \(y\): \(f(y+2)\).

Substitute and simplify

Substitute \(y + 2\) into the original function: \[ f(y+2) = 4(y+2)^3 - 11(y+2)^2 - 26(y+2) + 24 \] Expand each term and simplify.

Expand and Combine like terms

Expand each polynomial term: \[ (y+2)^3 = y^3 + 6y^2 + 12y + 8 \] \[ (y+2)^2 = y^2 + 4y + 4 \] Substitute these expansions back into \(f(y+2)\):\[ f(y+2) = 4(y^3 + 6y^2 + 12y + 8) - 11(y^2 + 4y + 4) - 26(y+2) + 24 \] Combine like terms after distributing: \[ f(y+2) = 4y^3 + 24y^2 + 48y + 32 - 11y^2 - 44y - 44 - 26y - 52 + 24 \] Simplify: \[ f(y+2) = 4y^3 + 13y^2 - 22y - 40 \]

Find the zeros of the simplified function

Set \( f(y+2) = 0 \) and solve for \( y \): \[ 4y^3 + 13y^2 - 22y - 40 = 0 \] Use factorization techniques (e.g., synthetic division, trial and error with rational root theorem) to find the values of y that satisfy the equation.

Solve for x

Once the roots of \(f(y)\) are found, where \( y = x - 2 \), solve for \( x \) by adding 2 back to each root found for \( y \).

Key concepts

cubic polynomials

A cubic polynomial is a type of polynomial where the highest degree of the variable is three. For example, in the given function, we have, \(f(x) = 4x^3 - 11x^2 - 26x + 24\). Here:

• \