Factor $$ 2 x^{2}+3 x-9 $$ $$ \begin{array}{l} ext { (2x+9)(x-1) } \ ext { Prime } \ ext { (x-3)(2x-3) } \ ext { Ox+3) }\end{array} $$

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Factor $$ 2 x^{2}+3 x-9 $$ $$ \begin{array}{l}	ext { (2x+9)(x-1) } \ 	ext { Prime } \ 	ext { (x-3)(2x-3) } \ 	ext { Ox+3) }\end{array} $$

UpStudy AI · Accepted Answer

1. We start with the quadratic expression:
   $$
   2x^2 + 3x - 9
   $$

2. To factor a quadratic of the form $$
   ax^2 + bx + c
   $$, we first compute the discriminant:
   $$
   \Delta = b^2 - 4ac = 3^2 - 4(2)(-9) = 9 + 72 = 81
   $$

3. Since $$\Delta = 81$$ is a perfect square, factorization over the integers is possible. The roots are given by:
   $$
   x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-3 \pm 9}{4}
   $$
   
   This yields:
   $$
   x = \frac{6}{4} = \frac{3}{2} \quad \text{and} \quad x = \frac{-12}{4} = -3
   $$

4. Writing the factors in terms of the roots, we have:
   $$
   x - \frac{3}{2} = 0 \quad \text{and} \quad x + 3 = 0
   $$

5. To eliminate the fraction in the first factor, multiply it by $$2$$:
   $$
   2\left(x - \frac{3}{2}\right) = 2x - 3
   $$
   
   Since we multiplied one factor by $$2$$ we must account for it by dividing the constant factor from the product (or simply note that the leading coefficient $$2$$ is used in $$2x-3$$). Thus, the complete factorization is:
   $$
   (2x - 3)(x + 3)
   $$

6. Therefore, the factorization of $$2x^2 + 3x - 9$$ is:
   $$
   (2x - 3)(x + 3)
   $$
$$2x^2 + 3x - 9$$ factors to $$(2x - 3)(x + 3)$$.

Factor \( 2 x^{2}+3 x-9 \) \( \begin{array}{l}\text { (2x+9)(x-1) } \\ \text { Prime } \\ \text { (x-3)(2x-3) } \\ \text { Ox+3) }\end{array} \)

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