Question

Factorise \(\nu^{3}-3 \nu^{2}-33 \nu+35\) given that \((\nu-7)\) is a factor.

Short answer

Question: Completely factorize the following polynomial: \(\nu^{3}-3 \nu^{2}-33 \nu+35\) given that \((\nu-7)\) is a factor. Answer: \(\nu^{3}-3 \nu^{2}-33 \nu+35 = (\nu-7)(\nu + 5)(\nu - 1)\).

Step-by-step solution

Perform synthetic division with the given factor

We are given that \((\nu-7)\) is a factor of the polynomial. We can use synthetic division to divide the polynomial by \((\nu-7)\). Here's how to set up the synthetic division: ``` 7 | 1 -3 -33 35 7 28 -35 ------------------- 1 4 -5 0 ``` The result of the synthetic division is a quadratic polynomial: \(\nu^{2} + 4 \nu - 5\). So, \(\nu^{3}-3 \nu^{2}-33 \nu+35 = (\nu-7)(\nu^{2} + 4 \nu - 5)\).

Factor the quadratic polynomial

Now, we need to factor the quadratic polynomial \(\nu^{2} + 4 \nu - 5\). We will do this by finding two numbers that multiply to \(-5\) and add to \(4\). The two numbers are \(5\) and \(-1\). Thus, we can factor the quadratic as follows: \(\nu^{2} + 4\nu -5 =(\nu + 5)(\nu - 1)\). Now the original polynomial is completely factored: \(\nu^{3}-3 \nu^{2}-33 \nu+35 = (\nu-7)(\nu + 5)(\nu - 1)\).

Key concepts

Synthetic Division

Synthetic division is a simplified way to divide a polynomial by a binomial of the form \((x - c)\). It’s particularly useful when you know one factor of the polynomial, making it quicker than long division. Let’s break down how it works by using the given example where \((u-7)\) is a factor:

• Write down the coefficients of the polynomial \(u^3 - 3u^2 - 33u + 35\). So, we use the numbers 1, -3, -33, 35.
• Place 7 to the left (since \((u - 7)\) is the divisor) of the 'synthetic division' bar.
• Bring down the leading coefficient 1.
• Multiply and add across the row: multiply 1 by 7, then add -3, getting 4. Continue this process through each coefficient.

The end result is \(1, 4, -5\). This corresponds to a new quadratic polynomial, \(u^2 + 4u - 5\). The last number should be 0 if the division is exact, confirming that \((u - 7)\) is a factor.

Quadratic Polynomial

Quadratic polynomials are expressions of the form \(ax^2 + bx + c\), where \(a, b,\) and \(c\) are constants. They form a parabola when plotted on a graph. In our exercise, the quadratic polynomial obtained from synthetic division is \(u^2 + 4u - 5\).

To factor this quadratic, we need numbers that multiply to \(-5\) and add to \(4\). After some trial and error, we find these numbers to be \(5\) and \(-1\).

• \(u^2 + 4u - 5 = (u + 5)(u - 1)\).
• This method makes use of the fact that the roots (or solutions) of the polynomial can help in factoring it, by setting each binomial expression equal to zero.

Quadratics can also be solved using the quadratic formula or by completing the square when they do not factor neatly.

Factor Theorem

The Factor Theorem is a special case of the Remainder Theorem, which provides a useful connection between polynomials and their roots. It states that if \(f(c) = 0\) for a polynomial \(f(x)\), then \((x-c)\) is a factor of \(f(x)\). Thus, determining that \((u - 7)\) is a factor meant checking that substituting \(u = 7\) zeros out our polynomial.

• We used synthetic division to quickly verify this in the exercise.
• The result revealed \((u - 7)\) as a factor without a remainder, supporting the Factor Theorem.

This foundational theorem is critical in factoring higher-degree polynomials, offering an insightful lens to understand roots and their related binomial factors.