Question

Expand the expression to a polynomial in standard form:
left parenthesis, x, squared, plus, x, minus, 2, right parenthesis, left parenthesis, x, squared, minus, x, minus, 6, right parenthesis
(x
2
+x−2)(x
2
−x−6)






Solve the problem just like the example below
⬇️
Expand the expression to a polynomial in standard form:
left parenthesis, x, squared, minus, x, minus, 2, right parenthesis, left parenthesis, minus, x, squared, plus, 3, x, minus, 6, right parenthesis
(x
2
−x−2)(−x
2
+3x−6)

minus, x, squared
−x
2
plus, 3, x
+3x
minus, 6
−6
x, squared
x
2
minus, x, to the power 4
−x
4
3, x, cubed
3x
3
minus, 6, x, squared
−6x
2
minus, x
−x
x, cubed
x
3
minus, 3, x, squared
−3x
2
6, x
6x
minus, 2
−2
2, x, squared
2x
2
minus, 6, x
−6x
12
12
Combine like terms:
Combine like terms:
minus, x, to the power 4 , plus, 4, x, cubed, minus, 7, x, squared, plus, 12
−x
4
+4x
3
−7x
2
+12

Answer 1

Sure! Here's the step-by-step expansion of the expression \((x^2 + x - 2)(x^2 - x - 6)\):

1. **Write down the expression:**
\[
(x^2 + x - 2)(x^2 - x - 6)
\]

2. **Distribute each term in the first polynomial to each term in the second polynomial:**

- Distributing \(x^2\):
\[
x^2 \cdot x^2 = x^4
\]
\[
x^2 \cdot (-x) = -x^3
\]
\[
x^2 \cdot (-6) = -6x^2
\]

- Distributing \(x\):
\[
x \cdot x^2 = x^3
\]
\[
x \cdot (-x) = -x^2
\]
\[
x \cdot (-6) = -6x
\]

- Distributing \(-2\):
\[
-2 \cdot x^2 = -2x^2
\]
\[
-2 \cdot (-x) = 2x
\]
\[
-2 \cdot (-6) = 12
\]

3. **Combine all the results:**
\[
x^4 - x^3 - 6x^2 + x^3 - x^2 - 6x - 2x^2 + 2x + 12
\]

4. **Combine like terms:**
- From the \(x^3\) terms: \(-x^3 + x^3 = 0\)
- From the \(x^2\) terms: \(-6x^2 - x^2 - 2x^2 = -9x^2\)
- From the \(x\) terms: \(-6x + 2x = -4x\)
- The constant term is \(12\).

5. **Write the final expression in standard form:**
\[
x^4 - 9x^2 - 4x + 12
\]

So the final expanded polynomial in standard form is:
\[
\boxed{x^4 - 9x^2 - 4x + 12}
\]