A quadratic equation $$ a x^{2}+b x+c=0 $$ has -6 and 3 as solutions. Find the values of $$ b $$ and $$ c $$ if the value of $$ a $$ is 1. (Hint: Use the zero-factor property in reverse.) The quadratic equation has $$ a=1, b=\square $$ and $$ c=\square $$

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To find the values of $$ b $$ and $$ c $$ in the quadratic equation $$ x^{2} + b x + c = 0 $$ with solutions $$ -6 $$ and $$ 3 $$, follow these steps:

### Step 1: Use the Zero-Factor Property
The zero-factor property states that if $$ r $$ and $$ s $$ are roots of the quadratic equation, then the equation can be written as:
$$
(x - r)(x - s) = 0
$$

Given the roots $$ -6 $$ and $$ 3 $$, substitute them into the equation:
$$
(x - (-6))(x - 3) = (x + 6)(x - 3)
$$

### Step 2: Expand the Equation
Expand the product to find the standard form of the quadratic equation:
$$
(x + 6)(x - 3) = x^{2} - 3x + 6x - 18 = x^{2} + 3x - 18
$$

### Step 3: Identify $$ b $$ and $$ c $$
Comparing the expanded equation to the standard form $$ x^{2} + b x + c = 0 $$:
$$
b = 3 \quad \text{and} \quad c = -18
$$

### Final Answer
When $$ a = 1 $$, the quadratic equation is:
$$
x^{2} + 3x - 18 = 0
$$
Thus,
$$
b = 3 \quad \text{and} \quad c = -18
$$
When $$ a = 1 $$, the quadratic equation is $$ x^{2} + 3x - 18 = 0 $$. So, $$ b = 3 $$ and $$ c = -18 $$.

A quadratic equation has -6 and 3 as solutions. Find the values of and if the value of is

(Hint: Use the zero-factor property in reverse.)
The quadratic equation has and

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