Question

1. if x = 3 is a root of $x^3 + 4x^2 - 9x - 36 = 0$, use synthetic division to factor the polynomial complete list all real solutions of the equation.

Explanation:

Step1: Set up synthetic division

We are dividing the polynomial \(x^{3}+4x^{2}-9x - 36\) by \((x - 3)\) (since \(x = 3\) is a root). The coefficients of the polynomial are \(1,4,-9,-36\).
We set up synthetic division as follows:

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Step2: Write the quotient polynomial

The result of the synthetic division gives us the quotient polynomial \(x^{2}+7x + 12\) (since the remainder is 0) and the factor \((x - 3)\). So the polynomial can be factored as \((x - 3)(x^{2}+7x + 12)\).

Step3: Factor the quadratic polynomial

We factor \(x^{2}+7x + 12\). We need two numbers that multiply to 12 and add up to 7. Those numbers are 3 and 4. So \(x^{2}+7x + 12=(x + 3)(x + 4)\).

Step4: Find all real solutions

Now we have the polynomial factored completely as \((x - 3)(x + 3)(x + 4)=0\).
To find the real solutions, we set each factor equal to zero:

• \(x - 3=0\) gives \(x = 3\)
• \(x + 3=0\) gives \(x=-3\)
• \(x + 4=0\) gives \(x=-4\)

Answer:

The real solutions of the equation \(x^{3}+4x^{2}-9x - 36 = 0\) are \(x=-4\), \(x=-3\), and \(x = 3\).