What is the sum of the polynomials (−x² + 9) + (−3x² − 11x + 4)?

Step-by-step solution

1. 1

   Drop the parentheses — addition needs no sign flip. Because the two polynomials are added, every term keeps the sign it already has:

   $-x^2 + 9 - 3x^2 - 11x + 4$

   This is the one thing that separates a sum from a difference, where the second bracket would have to be negated term by term.

2. 2

   Group the terms by degree. Like terms are terms with the identical power of $x$, and only those may be combined:

   $(-x^2 - 3x^2) + (-11x) + (9 + 4)$

   The $x$ column has a single entry, since the first polynomial has no linear term.

3. 3

   Combine each column. Add the coefficients within each group, keeping the power unchanged:

   $-x^2 - 3x^2 = -4x^2, \qquad 9 + 4 = 13$

   The linear term stays $-11x$ untouched. Note $-1 - 3 = -4$, not $-2$: the leading coefficient of $-x^2$ is $-1$, not $-3$.

4. 4

   Write the result in standard form. Ordering from highest degree to lowest:

   $-4x^2 - 11x + 13$

5. 5

   Check numerically at a convenient value. Substituting $x = 1$ into the original expression gives

   $(-1 + 9) + (-3 - 11 + 4) = 8 + (-10) = -2$

   and into the answer

   $-4 - 11 + 13 = -2 \ \checkmark$

   Testing a second value such as $x = 2$ gives $5 + (-30) = -25$ from the original and $-16 - 22 + 13 = -25$ from the answer, which is enough to confirm the coefficients of a quadratic.