- Divide the given polynomial \(f(x) = x^3 - 7x^2 - 16x + 112\) by the known factor \((x-4)\) to get the quadratic factor \(x^2 - 3x - 28\).
- Factor the quadratic \(x^2 - 3x - 28\) into \((x-7)(x+4)\).
- Combine all factors to get \((x-7)(x-4)(x+4)\).
- Order the factors by their constant terms from smallest to largest to get \((x-7)(x-4)(x+4)\).
- The final answer is \(\boxed{(x-7)(x-4)(x+4)}\).
Explanation
- Analyzing the polynomial and the given root
The polynomial is given as \(f(x) = x^3 - 7x^2 - 16x + 112\). We are told that \(x=4\) is a root (or zero) of this function. By the Factor Theorem, if \(x=r\) is a root of a polynomial, then \((x-r)\) is a factor of that polynomial. Thus, \((x-4)\) is a factor of \(f(x)\).
- Performing polynomial division
To find the remaining factors, we divide the original polynomial \(f(x)\) by \((x-4)\). Performing polynomial long division (or synthetic division) of \(x^3 - 7x^2 - 16x + 112\) by \(x-4\):
- Dividing \(x^3\) by \(x\) gives \(x^2\).
- Multiplying \((x-4)\) by \(x^2\) gives \(x^3 - 4x^2\).
- Subtracting this from \(f(x)\) leaves \(-3x^2 - 16x + 112\).
- Dividing \(-3x^2\) by \(x\) gives \(-3x\).
- Multiplying \((x-4)\) by \(-3x\) gives \(-3x^2 + 12x\).
- Subtracting this from the remaining part leaves \(-28x + 112\).
- Dividing \(-28x\) by \(x\) gives \(-28\).
- Multiplying \((x-4)\) by \(-28\) gives \(-28x + 112\).
The remainder is \(0\), and the quotient is \(x^2 - 3x - 28\). So, \(f(x) = (x-4)(x^2 - 3x - 28)\).
- Factoring the quadratic quotient
Now we factor the quadratic polynomial \(x^2 - 3x - 28\). We look for two numbers that multiply to \(-28\) and add to \(-3\). The numbers \(-7\) and \(4\) satisfy these conditions: \(-7 \times 4 = -28\) and \(-7 + 4 = -3\).
Therefore, \(x^2 - 3x - 28 = (x-7)(x+4)\).
Combining this with the factor we found earlier, the fully factored form is \(f(x) = (x-4)(x-7)(x+4)\).
- Ordering the factors and concluding
To express the factors in the form \((x-a)(x-b)(x-c)\) where the constant terms are ordered from smallest to largest, we examine the constant terms of our factors: \(-4, -7, 4\).
Ordered, these are \(-7, -4, 4\).
Therefore, the factors written as \((x-a)(x-b)(x-c)\) where \(a \leq b \leq c\) are \((x-(-4))(x-4)(x-7)\) which is \((x+4)(x-4)(x-7)\). Wait, the constant terms are \(-7, -4, 4\).
Our factors are \((x-7)\), \((x+4)\), \((x-4)\).
Looking at the constant terms:
For \((x-7)\), constant is \(-7\).
For \((x-4)\), constant is \(-4\).
For \((x+4)\), constant is \(4\).
Ordering these: \(-7 \leq -4 \leq 4\).
The factored form is \((x-7)(x-4)(x+4)\).
Examples
Factoring polynomials is a fundamental skill in algebra with practical applications in engineering and physics. For example, when analyzing the motion of an object, you might arrive at a polynomial equation representing the object's height at different times. Factoring the polynomial helps you identify the specific times (zeros) when the object is at ground level, allowing you to calculate flight time and projectile path efficiently. Just as we broke down the cubic polynomial into its linear components, engineers break down complex physical systems into simpler, manageable parts to predict behavior and design better structures.