Problem 80 - Algebra and Trigonometry Real Mathematics, Real People

Question

Factor the polynomial completely. \(6 x^{3}-11 x^{2}-10 x\)

Step-by-Step Solution

Answer

The factored form of the polynomial \(6 x^{3}-11 x^{2}-10 x\) is \(x*(2x - 5)*(3x + 2)\).

Step 1: Write out the Polynomial

The polynomial given is \(6 x^{3}-11 x^{2}-10 x\).

Step 2: Factor Out a Common Term

Notice that all of the terms in the polynomial have \(x\) in them. Factor \(x\) from each term: \(x(6 x^{2}-11 x-10)\).

Step 3: Factor the Quadratic

Now, we need to factor the quadratic \(6 x^{2}-11 x-10\). Find two numbers that multiply to \(-60\) (which is \(6 * -10\)) and add to \(-11\). These numbers are \(-15\) and \(4\). Therefore, the factored form of the quadratic is \((2x - 5)(3x + 2)\).

Step 4: Write the Final Answer

Plugging \((2x - 5)\) and \((3x + 2)\) back to the factored x from Step 2, the final factored form of the polynomial is \(x*(2x - 5)*(3x + 2)\).

Key Concepts

Common Factor

When you start factoring a polynomial, always look for a common factor among its terms. Identifying this common factor simplifies the polynomial by reducing each term by that factor.
For the polynomial \(6x^3 - 11x^2 - 10x\), the common factor is \(x\). Each term in the polynomial contains \(x\), which can be taken out: \(x(6x^2 - 11x - 10)\).
This step is crucial because it reduces the complexity of the polynomial, making it easier to factor further. The goal is to simplify each term so that they form a more recognizable pattern for further steps. This "factoring out" process is the foundation for more complex polynomial operations.

Quadratic Factoring

Factoring a quadratic expression in the form \(ax^2 + bx + c\) involves finding two numbers that multiply to \(ac\) and add to \(b\). This step can be counter-intuitive, but it follows a logical approach.
In our current expression, \(6x^2 - 11x - 10\), the product \(ac\) is \(-60\) (since \(6 \times -10 = -60\)) and the sum \(b\) is \(-11\). We need to find two numbers, \(-15\) and \(4\), that satisfy these conditions.
These numbers break down the middle term \(-11x\) into \(-15x + 4x\), thus converting the quadratic into \(6x^2 - 15x + 4x - 10\). Grouping and factoring in pairs gives \((2x - 5)(3x + 2)\). This step showcases the beauty of quadratic factoring where a seemingly complicated expression breaks down into simpler binomial factors.

Multiplying Binomials

Once binomials are identified through factoring, understanding how to multiply them is key. Multiplying binomials, often referred to as using the FOIL method, strengthens the grasp on how these expressions interact.
Let's illustrate this with our factors, \((2x - 5)(3x + 2)\). When expanded, each term of the first binomial is multiplied with each term of the second one:

• **First:** \(2x \cdot 3x = 6x^2\)
  
• **Outside:** \(2x \cdot 2 = 4x\)
  
• **Inside:** \(-5 \cdot 3x = -15x\)
  
• **Last:** \(-5 \cdot 2 = -10\)
  

Combining these gives us the original quadratic \(6x^2 - 11x - 10\).
This multiplication reinforces the relationship between the expanded and factored form, confirming the correctness of the factorization process. Understanding this connection helps in reverse engineering, making factoring practices more intuitive for similar problems.