Question

Factor t^3 + 2t^2 - 3. Show your work

Answer 1

To factor the polynomial \( t^3 + 2t^2 - 3 \), we can use the Rational Root Theorem to test for possible rational roots, and then perform polynomial long division if necessary.

### Step 1: Apply the Rational Root Theorem

The Rational Root Theorem states that any potential rational root of the polynomial \( p(t) = a_n t^n + a_{n-1} t^{n-1} + ... + a_1 t + a_0 \) is in the form \( \frac{p}{q} \), where \( p \) divides the constant term \( a_0 \) and \( q \) divides the leading coefficient \( a_n \).

For \( t^3 + 2t^2 - 3 \):
- The constant term \( a_0 = -3 \).
- The leading coefficient \( a_n = 1 \).

The factors of \( -3 \) are \( \pm 1, \pm 3 \). The factors of \( 1 \) are \( \pm 1 \).

Thus, the possible rational roots are:
- \( \pm 1 \)
- \( \pm 3 \)

### Step 2: Test the Possible Roots

We will test these possible roots by substituting them into the polynomial.

1. Testing \( t = 1 \):
\[
p(1) = 1^3 + 2(1)^2 - 3 = 1 + 2 - 3 = 0
\]
Since \( p(1) = 0 \), \( t = 1 \) is a root.

### Step 3: Polynomial Long Division

Now that we know \( t - 1 \) is a factor, we can divide the polynomial \( t^3 + 2t^2 - 3 \) by \( t - 1 \).

1. Set up the division:
\[
\begin{array}{r|rrr}
& 1 & 2 & 0 & -3 \\
\hline
1 & 1 & 1 & 1 \\
\hline
& 1 & 3 & 1 \\
\end{array}
\]

2. Perform the division:
- Divide the first term \( t^3 \) by \( t \) to get \( t^2 \).
- Multiply \( t^2 \) by \( t - 1 \) to get \( t^3 - t^2 \).
- Subtract that from the original polynomial:
\[
(t^3 + 2t^2) - (t^3 - t^2) = 3t^2
\]
- Bring down the \( -3 \):
\[
3t^2 + 0 - 3
\]
- Repeat the process:
- Divide \( 3t^2 \) by \( t \) to get \( 3t \).
- Multiply \( 3t \) by \( t - 1 \) to get \( 3t^2 - 3t \).
- Subtract:
\[
(3t^2) - (3t^2 - 3t) = 3t
\]
- Bring down the last term \( -3 \):
\[
3t - 3
\]
- Divide \( 3t \) by \( t \) to get \( 3 \).
- Multiply \( 3 \) by \( t - 1 \) to get \( 3t - 3 \).
- Subtract:
\[
(3t) - (3t - 3) = 0
\]

The division gives:
\[
t^3 + 2t^2 - 3 = (t - 1)(t^2 + 3t + 3)
\]

### Step 4: Factor Further (if possible)

Now we need to check if \( t^2 + 3t + 3 \) can be factored further. To do this, we can compute the discriminant:
\[
D = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot 3 = 9 - 12 = -3
\]
Since the discriminant is negative, \( t^2 + 3t + 3 \) does not have real roots and cannot be factored further over the real numbers.

### Final Answer

Thus, the complete factorization of the polynomial \( t^3 + 2t^2 - 3 \) is:
\[
\boxed{(t - 1)(t^2 + 3t + 3)}
\]