Quadratics are \( 2^{nd} \) degree polynomials.

One way to determine the roots of a quadratic equation (or the zeros of a quadratic function) is to factor the quadratic.

For example, to find the zeros (or \(x\)-intercepts) of the given parabola, we would set \(y = 0\) and factor.

Therefore, \( x = \frac{5}{2} \) and \( x = -4 \) are the roots, or solutions, to the equation \( 0 = 2x^2 + 3x - 20 \),

and \( x = \frac{5}{2} \) and \( x = -4 \) are the zeros, or \( x \)-intercepts, of the function \( y = 2x^2 + 3x - 20 \).

State the roots of the equation, $\mathit{x}\in \mathit{R}$.

((((((((b)*(r))*(a))*(c))*(_))*(p))*1.0)*(_))*1.02.718282x ((s)*(g))*1.0 f((((((((b)*(r))*(a))*(c))*(_))*(p))*1.0)*(_))*2.0^{(((p)*(o))*(w))*1.0}(ax ((s)*(g))*2.0 b)(cx ((s)*(g))*3.0 d)^{(((p)*(o))*(w))*3.0} = 0

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Note :

1. Roots with multiplicity greater than 1 should only be entered once.
2. Provide exact values only (explained).
3. Separate answers with a comma (e.g. 8,-2/3,0,sqrt(3)). Order does not matter

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Solve \(2x^3+8x^2-3x-12=0\).

Solution

The first step in solving polynomial equations of degree \(2\) or higher is to ensure that all terms are to one side of the equation, leaving \(0\) on the other side.

Once factored, we can then rely on the premise that the product of factors is equal to zero when one of the factors is zero.

Since \((2)(-12)=(8)(-3)\), this cubic can be factored by grouping.

Thus, \( x + 4 = 0 \) or \( 2x^2 - 3 = 0 \).

Solve \(2x^3+8x^2-3x-12=0\).

Solution

Although the radical answer is correct, good mathematical practice is to remove the radical values from the denominator. This is called rationalizing the denominator.

We rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{2}\).

Therefore, the roots are \( x = -4\) and \( \pm \dfrac{\sqrt{6}}{2} \).