- The monic property identifies the leading coefficient as 1, and the double root at \(x=4\) identifies the factors as \((x-4)^2\).
- Expanding \((x-4)^2\) using the binomial identity \(u^2 - 2uv + v^2\) yields \(x^2 - 8x + 16\).
- The correct polynomial is \(y = x^2 - 8x + 16\).
- Final answer: \(\boxed{y=x^2-8x+16}\)
Explanation
- Understanding the Problem and Factored Form
We are given a quadratic polynomial \(p(x) = ax^2 + bx + c\) that is 'monic', meaning its leading coefficient \(a\) is 1. We are told that this polynomial has a single zero at \(x = 4\), which is a double root. According to the factor theorem, if \(r\) is a root of a polynomial, then \((x - r)\) is a factor. Since \(x = 4\) is a double root, \((x - 4)\) must be a factor twice. Therefore, the polynomial can be written in its factored form as \(p(x) = 1 \cdot (x - 4)^2\).
- Expanding to Standard Form
To express the polynomial in standard form \(p(x) = ax^2 + bx + c\), we need to expand the expression \((x - 4)^2\). Using the algebraic identity for the square of a binomial, \((u - v)^2 = u^2 - 2uv + v^2\), we let \(u = x\) and \(v = 4\). This gives us:
\[(x - 4)^2 = x^2 - 2(x)(4) + 4^2 = x^2 - 8x + 16.\]
Thus, the monic quadratic polynomial in standard form is \(y = x^2 - 8x + 16\).
- Selecting the Correct Option
Comparing our result \(y = x^2 - 8x + 16\) with the given choices:
- \(y = x^2 - 4x + 16\)
- \(y = x^2 + 8x + 16\)
- \(y = x^2 + 16\)
- \(y = x^2 - 8x + 16\)
Our calculated polynomial matches the last option. Therefore, the correct monic quadratic is \(y = x^2 - 8x + 16\).
Examples
In engineering and physics, understanding the behavior of quadratic functions is essential, particularly when analyzing systems with equilibrium points. A 'double root' often corresponds to a critical point where the function touches the x-axis but doesn't cross it, representing a state of perfect balance or a stationary point in a trajectory. For example, if you are modeling the height of a projectile that perfectly grazes a target at height zero, the quadratic equation representing its path might have a double root. Mastering how to construct these equations allows you to predict such physical behaviors accurately.