Question

\left(2x^2+x+3\right)\left(3x^2-2x+6\right)

you made me get the other question wrong so dont make me fail

Answer 1

To multiply the two polynomials \((2x^2 + x + 3)(3x^2 - 2x + 6)\), we'll use the distributive property (also known as the FOIL method for binomials) to ensure we multiply each term in the first polynomial by each term in the second polynomial.

Let's perform the multiplication step by step:

1. **Distribute** \(2x^2\):
- \(2x^2 \cdot 3x^2 = 6x^4\)
- \(2x^2 \cdot (-2x) = -4x^3\)
- \(2x^2 \cdot 6 = 12x^2\)

2. **Distribute** \(x\):
- \(x \cdot 3x^2 = 3x^3\)
- \(x \cdot (-2x) = -2x^2\)
- \(x \cdot 6 = 6x\)

3. **Distribute** \(3\):
- \(3 \cdot 3x^2 = 9x^2\)
- \(3 \cdot (-2x) = -6x\)
- \(3 \cdot 6 = 18\)

Now, combine all these results together:
\[
6x^4 + (-4x^3 + 3x^3) + (12x^2 - 2x^2 + 9x^2) + (6x - 6x) + 18
\]

Now, simplify by combining like terms:
- For \(x^4\): \(6x^4\)
- For \(x^3\): \(-4x^3 + 3x^3 = -1x^3\)
- For \(x^2\): \(12x^2 - 2x^2 + 9x^2 = 19x^2\)
- For \(x\): \(6x - 6x = 0\)
- Constant term: \(18\)

Thus, the final result is:
\[
6x^4 - x^3 + 19x^2 + 18
\]

So, \((2x^2 + x + 3)(3x^2 - 2x + 6) = 6x^4 - x^3 + 19x^2 + 18\).