Question

Factorize \[ 2\cot^2\theta-\cot\theta-3 \]

• A. \((2\cot\theta-3)(\cot\theta+1)\)
• B. \((2\cot\theta-1)(\cot\theta+3)\)
• C. \((2\cot\theta+3)(\cot\theta-1)\)
• D. \((2\cot\theta+1)(\cot\theta-3)\)

Hint:

For quadratic expressions of the form \[ ax^2+bx+c, \] find two numbers whose product is \(ac\) and whose sum is \(b\). Then use factorization by grouping.

Answer 1

The Correct Option is A

Step 1: Treat \(\cot\theta\) as a variable and factor directly.
Let \(u = \cot\theta\). Factor \(2u^2 - u - 3\): find two numbers with product \(2(-3)=-6\) and sum \(-1\): these are \(-3\) and \(2\). So \(2u^2 - u - 3 = 2u^2 - 3u + 2u - 3 = u(2u-3)+1(2u-3) = (2u-3)(u+1)\).

Step 2: Substitute back.
\[(2\cot\theta - 3)(\cot\theta + 1).\]
\[\boxed{(2\cot\theta - 3)(\cot\theta + 1)}\]