Simplify, giving your answer in factors : `3(x-1)^(2) + 5(x-1)(x+4)-2(x+4)^(2)`. The answer is

Options

A. `(2x+7)(3x-7)`
B. `(x-7)(2x+7)`
C. `(2x-3)(x+7)`
D. `(2x-7)(3x+7)`

Answer

To simplify the expression \(3(x-1)^{2} + 5(x-1)(x+4) - 2(x+4)^{2}\) and give the answer in factors, follow these steps:

### Step 1: Substitute Variables
Let \(y = x - 1\) and \(z = x + 4\).

### Step 2: Rewrite the Expression
Now, rewrite the expression using \(y\) and \(z\):
\[ 3y^{2} + 5y(z) - 2z^{2} \]

### Step 3: Substitute \(z\) in Terms of \(y\)
Since \(z = x + 4\) and \(y = x - 1\), we can express \(z\) in terms of \(y\):
\[ z = y + 5 \]
Now substitute \(z\) into the expression:
\[ 3y^{2} + 5y(y + 5) - 2(y + 5)^{2} \]

### Step 4: Expand the Expression
Now expand the expression:
\[ 3y^{2} + 5y^{2} + 25y - 2(y^{2} + 10y + 25) \]
\[ = 3y^{2} + 5y^{2} + 25y - 2y^{2} - 20y - 50 \]
\[ = (3y^{2} + 5y^{2} - 2y^{2}) + (25y - 20y) - 50 \]
\[ = 6y^{2} + 5y - 50 \]

### Step 5: Factor the Quadratic Expression
Now, we need to factor \(6y^{2} + 5y - 50\). We look for two numbers that multiply to \(6 \times -50 = -300\) and add to \(5\). The numbers \(20\) and \(-15\) work:
\[ 6y^{2} + 20y - 15y - 50 \]
Group the terms:
\[ = (6y^{2} + 20y) + (-15y - 50) \]
Factor out common terms:
\[ = 2y(3y + 10) - 5(3y + 10) \]
Now factor by grouping:
\[ = (2y - 5)(3y + 10) \]

### Step 6: Substitute Back for \(y\)
Now substitute back \(y = x - 1\):
\[ = (2(x - 1) - 5)(3(x - 1) + 10) \]
\[ = (2x - 2 - 5)(3x - 3 + 10) \]
\[ = (2x - 7)(3x + 7) \]

### Final Answer
Thus, the factorized form of the given expression is:
\[ (2x - 7)(3x + 7) \]