Question

the movement of the progress bar may be uneven because questions can be worth more or less (factor: $6x^2 - 7x + 2$ $\bigcirc$ $(2x + 1)(3x - 2)$ $\bigcirc$ $(2x - 1)(3x - 2)$ $\bigcirc$ $2x - 1 + 3x - 2$ $\bigcirc$ $(3x - 2)(4x - 2)$

Explanation:

Step1: Recall factoring quadratic

We need to factor \(6x^2 - 7x + 2\). For a quadratic \(ax^2+bx + c\), we look for two numbers that multiply to \(a\times c = 6\times2=12\) and add to \(b=-7\) (note the signs, since \(c\) is positive and \(b\) is negative, both numbers are negative). The numbers are \(-3\) and \(-4\) (since \(-3\times -4 = 12\) and \(-3 + (-4)=-7\)).

Step2: Rewrite middle term

Rewrite \(-7x\) as \(-3x-4x\): \(6x^2-3x - 4x + 2\).

Step3: Group and factor

Group the first two and last two terms: \((6x^2 - 3x)+(-4x + 2)\). Factor out \(3x\) from the first group and \(-2\) from the second group: \(3x(2x - 1)-2(2x - 1)\). Now factor out \((2x - 1)\): \((2x - 1)(3x - 2)\). We can also check by expanding the options:

• Expand \((2x + 1)(3x - 2)=6x^2-4x+3x - 2=6x^2 - x - 2\) (not correct).
• Expand \((2x - 1)(3x - 2)=6x^2-4x-3x + 2=6x^2 - 7x + 2\) (correct).
• \(2x - 1+3x - 2=5x - 3\) (not a product, so wrong).
• Expand \((3x - 2)(4x - 2)=12x^2-6x-8x + 4=12x^2 - 14x + 4\) (not correct).

Answer:

B. \((2x - 1)(3x - 2)\) (assuming the second option is labeled B, as per the order: first option A, second B, etc. If the options are labeled with the given expressions, the correct one is \((2x - 1)(3x - 2)\))