Which expression in factored form is equivalent to this expression?



$4\left(x^2-2 x\right)-2\left(x^2-3\right)$


A$(2 x-3)(x+1)$
B$(2 x+3)(x+1)$
C$2(x+1)(x+3)$
D$2(x-1)(x-3)$

Verified Answer :

  • Expand the expression: \(4(x^2-2x) - 2(x^2-3) = 4x^2 - 8x - 2x^2 + 6\).
  • Combine like terms to get the simplified quadratic: \(2x^2 - 8x + 6\).
  • Factor out the common factor of 2: \(2(x^2 - 4x + 3)\).
  • Factor the trinomial \(x^2 - 4x + 3\) into \((x-1)(x-3)\) to reach the final form \(\boxed{2(x-1)(x-3)}\).

Explanation

  1. Analysis of the Expression

The problem asks us to find the factored form of the expression \(4(x^2-2x) - 2(x^2-3)\). To achieve this, we will first expand the expression by distributing the constants, combine the resulting like terms, and then factor the resulting quadratic expression.

  1. Expanding and Simplifying the Expression

First, we distribute the constants across the terms inside the parentheses:
\[4(x^2-2x) - 2(x^2-3) = (4 \times x^2 - 4 \times 2x) - (2 \times x^2 - 2 \times 3)\]
\[= 4x^2 - 8x - 2x^2 + 6\]
Next, we group the like terms (the \(x^2\) terms, the \(x\) terms, and the constant terms):
\[(4x^2 - 2x^2) - 8x + 6 = 2x^2 - 8x + 6\]
The expanded and simplified expression is \(2x^2 - 8x + 6\).

  1. Factoring the Quadratic Expression

Now, we factor the simplified quadratic expression \(2x^2 - 8x + 6\). We notice that each coefficient is divisible by 2, so we can factor out the greatest common factor:
\[2x^2 - 8x + 6 = 2(x^2 - 4x + 3)\]
To factor the quadratic \(x^2 - 4x + 3\), we look for two numbers that multiply to 3 and add up to -4. These two numbers are -1 and -3. Thus, we can write the expression as:
\[2(x-1)(x-3)\]
Comparing this result with the given choices, we see that it corresponds to the final option provided in the list.

Examples

Factoring algebraic expressions is a fundamental skill used in physics and engineering to simplify equations of motion or circuit analysis. For instance, if you have a projectile motion equation represented by a quadratic, factoring it allows you to easily find the roots, which represent the times when the projectile hits the ground. Understanding these techniques helps in solving for unknowns efficiently and modeling real-world phenomena accurately.