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Fill in the missing pieces in the factorization of x^3+2x^2−9x−18

(x^2-__)(x+__)

1 answer

To factor the polynomial \( x^3 + 2x^2 - 9x - 18 \) in the form \( (x^2 - A)(x + B) \), we start by expanding the expression:

\[ (x^2 - A)(x + B) = x^3 + Bx^2 - Ax^2 - AB \]

This simplifies to:

\[ x^3 + (B - A)x^2 - AB \]

Next, we want to match coefficients with the polynomial \( x^3 + 2x^2 - 9x - 18 \).

From the polynomial, we have:

  • The coefficient of \( x^2 \): \( B - A = 2 \)
  • The coefficient of \( x \): \( -AB = -9 \)
  • The constant term: \( 0 = -18 \) (which will be handled later)

We can rewrite this as two equations:

  1. \( B - A = 2 \) (1)
  2. \( AB = 9 \) (2)

From equation (1), we can express \( B \) in terms of \( A \):

\[ B = A + 2 \]

Substituting \( B \) from equation (1) into equation (2), we have:

\[ A(A + 2) = 9 \]

Expanding this gives us:

\[ A^2 + 2A - 9 = 0 \]

Next, we can solve this quadratic equation using the quadratic formula:

\[ A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-9)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 36}}{2} = \frac{-2 \pm \sqrt{40}}{2} = \frac{-2 \pm 2\sqrt{10}}{2} = -1 \pm \sqrt{10} \]

Thus, we get two possible values for \( A \):

  1. \( A = -1 + \sqrt{10} \)
  2. \( A = -1 - \sqrt{10} \)

Now calculating \( B \):

For \( A = -1 + \sqrt{10} \):

\[ B = (-1 + \sqrt{10}) + 2 = 1 + \sqrt{10} \]

For \( A = -1 - \sqrt{10} \):

\[ B = (-1 - \sqrt{10}) + 2 = 1 - \sqrt{10} \]

Now, in the factorization, we recognize that \( A \) will usually be a simple number, so we can try trial and error with the potential simple integers.

Using the Rational Root Theorem or simply testing integer factors of \( -18 \) along with \( \text{gcd}(1, 1) \), let's test \( x = -3 \):

Plugging into \( x^3 + 2x^2 - 9x - 18 \):

  1. Calculate \( (-3)^3 + 2(-3)^2 - 9(-3) - 18 \) = \( -27 + 18 + 27 - 18 = 0\).

Thus \( x = -3 \) is a root. Therefore, we can factor \( x + 3 \) out:

Performing polynomial long division of \( x^3 + 2x^2 - 9x - 18 \) by \( x + 3 \):

  1. Divide the leading term: \( x^3 \div x = x^2 \)
  2. Multiply \( x^2(x + 3) = x^3 + 3x^2 \), and subtract from the original polynomial (\( x^3 + 2x^2 - 9x - 18 \)):
    • \( (2x^2 - 3x^2) = -x^2\)
    • \( (-9x) - 0 = -9x \)
    • Bringing down \(-18\) gives us \( -x^2 - 9x - 18 \).
  3. Divide \( -x^2 \div x = -x \)
  4. Multiply and subtract:
  5. Notice the resulting constant leads to \( -(-3)(-x - 3) = -9\).
  6. This final yields zero remainder confirming \( x + 3 \) fully divides.

Thus, we find \( x^3 + 2x^2 - 9x - 18 = (x + 3)(x^2 - x - 6)\). Factoring \(x^2 - x - 6 = (x - 3)(x + 2)\),

Thus:

\[ x^3 + 2x^2 - 9x - 18 = (x + 3)(x - 3)(x + 2) \]

Final answer fitting into form:

\[ = (x^2 - 3)(x + 3) \text{ fits format } (x^2 - A)(x + B), A = 3, \boxed{3}, B = 3 \] leading to (\(A=3, B=3\)), so completing your format { (x^2-3)(x+3) with missing pieces:

So, the answer is: (x^2 - 3)(x + 3)