factor the following trinomial using substitution 14c to the 4th power - 33csquared + 18

To factor the trinomial \( 14c^4 - 33c^2 + 18 \) using substitution, follow these steps:

### Step 1: Substitution
Let’s substitute \( c^2 \) with a new variable \( x \). So, we define:

\[
x = c^2
\]

Now the given trinomial becomes:

\[
14x^2 - 33x + 18
\]

### Step 2: Factor the quadratic expression
Next, we factor the quadratic \( 14x^2 - 33x + 18 \) using the method of factoring by grouping.

- Multiply the coefficient of \( x^2 \) (14) by the constant term (18): 

\[
14 \times 18 = 252
\]

- Now find two numbers that multiply to 252 and add up to -33. These numbers are -21 and -12.

So, we can split the middle term:

\[
14x^2 - 21x - 12x + 18
\]

### Step 3: Group terms and factor by grouping
Group the terms as follows:

\[
(14x^2 - 21x) - (12x - 18)
\]

Factor out the greatest common factor (GCF) from each group:

\[
7x(2x - 3) - 6(2x - 3)
\]

Now, factor out the common binomial factor \( (2x - 3) \):

\[
(7x - 6)(2x - 3)
\]

### Step 4: Substitute back \( x = c^2 \)
Now, substitute \( x = c^2 \) back into the factors:

\[
(7c^2 - 6)(2c^2 - 3)
\]

Thus, the factored form of the trinomial \( 14c^4 - 33c^2 + 18 \) is:

\[
(7c^2 - 6)(2c^2 - 3)
\]

Would you like more details or further explanation on any step?

Here are 5 related questions:
1. How would you factor a trinomial with a negative leading coefficient?
2. Can you solve for \( c \) from the equation \( (7c^2 - 6)(2c^2 - 3) = 0 \)?
3. How does substitution simplify factoring higher-degree polynomials?
4. What is the difference between factoring by grouping and other factoring methods?
5. What happens when the trinomial cannot be factored easily?

**Tip**: When factoring, always check if common factors can be factored out first, to simplify the expression!

Learn how to factor the trinomial 14c^4 - 33c^2 + 18 using substitution. This problem covers key algebraic concepts like substitution and factoring by grouping. Suitable for students in Grades 9-11.

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