4x^4-7x^2-2   completely factor the polynomial

To completely factor the polynomial \(4x^4 - 7x^2 - 2\), follow these steps:

### Step 1: Recognize the substitution.
Notice that the polynomial is in the form of a quadratic in terms of \(x^2\). Let's make a substitution:
\[
y = x^2
\]
This transforms the polynomial into:
\[
4y^2 - 7y - 2
\]

### Step 2: Factor the quadratic expression.
Now, factor the quadratic expression \(4y^2 - 7y - 2\).

We need to find two numbers that multiply to \(4 \times (-2) = -8\) and add up to \(-7\).

These numbers are \(-8\) and \(1\).

So, rewrite the middle term:
\[
4y^2 - 8y + y - 2
\]

### Step 3: Factor by grouping.
Now, factor by grouping:
\[
(4y^2 - 8y) + (y - 2)
\]
Factor out the common factors from each group:
\[
4y(y - 2) + 1(y - 2)
\]
Factor out the common binomial \(y - 2\):
\[
(4y + 1)(y - 2)
\]

### Step 4: Substitute back \(y = x^2\).
Replace \(y\) with \(x^2\) to get the factors in terms of \(x\):
\[
(4x^2 + 1)(x^2 - 2)
\]

### Final Answer
The completely factored form of the polynomial is:
\[
(4x^2 + 1)(x^2 - 2)
\]

Would you like more details on any of these steps, or do you have any other questions?

Here are 5 related questions to consider:
1. How would you factor \(x^4 - 5x^2 + 6\) using a similar substitution method?
2. Can you factor the expression \(4x^2 + 1\) further over the real numbers?
3. How would you solve the equation \(4x^4 - 7x^2 - 2 = 0\) after factoring?
4. What are the roots of the polynomial \(x^2 - 2 = 0\)?
5. How would you factor a cubic polynomial, like \(x^3 - 6x^2 + 11x - 6\)?

**Tip:** When dealing with higher-degree polynomials, substitution can simplify the problem by reducing it to a more familiar quadratic form.

Learn how to completely factor the polynomial 4x^4 - 7x^2 - 2 with a step-by-step solution. This problem involves factoring a polynomial of degree 4 using substitution and quadratic factoring techniques. Suitable for students in Grades 10-12.

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