Factorise `4x ^(4) + 3x ^(2) +1`

Options

A. `(2x ^(2) + x +1) (2x ^(2) +x -1)`
B. `(2x ^(2) + x +1) (2x ^(2) -x +1)`
C. `(2x ^(2) + x -1) (2x ^(2) -x +1)`
D. `(2x ^(2) - x +1) (2x ^(2) -x -1)`

Answer

To factorize the expression \(4x^4 + 3x^2 + 1\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 4x^4 + 3x^2 + 1 \]
Notice that this is a bi-quadratic expression, which can be treated as a quadratic in terms of \(x^2\).

### Step 2: Substitute \(y = x^2\)
Let \(y = x^2\). Then, we can rewrite the expression as:
\[ 4y^2 + 3y + 1 \]

### Step 3: Factor the quadratic
Next, we need to factor the quadratic \(4y^2 + 3y + 1\). We look for two numbers that multiply to \(4 \cdot 1 = 4\) and add up to \(3\). The numbers \(1\) and \(2\) fit this requirement.

### Step 4: Rewrite the middle term
We can rewrite \(3y\) as \(2y + 1y\):
\[ 4y^2 + 2y + 1y + 1 \]

### Step 5: Group the terms
Now, we group the terms:
\[ (4y^2 + 2y) + (1y + 1) \]

### Step 6: Factor by grouping
Now, factor out the common factors from each group:
\[ 2y(2y + 1) + 1(2y + 1) \]

### Step 7: Factor out the common binomial
Now, we can factor out the common binomial \((2y + 1)\):
\[ (2y + 1)(2y + 1) = (2y + 1)^2 \]

### Step 8: Substitute back \(y = x^2\)
Now, substitute back \(y = x^2\):
\[ (2x^2 + 1)^2 \]

### Final Answer
Thus, the factorization of the original expression \(4x^4 + 3x^2 + 1\) is:
\[ (2x^2 + 1)^2 \]

---