Question

1. write a possible equation for a polynomial whose graph has the horizontal intercepts: x = 2, -1/2, and -3. then, re - write it in standard form.
2. looking at the zeros, what is the function of the graph shown here? write it in standard form.
3. evaluate 6(x - 2)(x - 3)+4(x - 2)(x - 5) when x = -3.

Explanation:

Step1: Write polynomial from roots

If \(x = a\) is a root of a polynomial, then \((x - a)\) is a factor. For roots \(x = 2,-\frac{1}{2},- 3\), the polynomial \(P(x)=k(x - 2)(x+\frac{1}{2})(x + 3)\). Let \(k = 1\), so \(P(x)=(x - 2)(x+\frac{1}{2})(x + 3)\).

Step2: Multiply first two factors

\((x - 2)(x+\frac{1}{2})=x^{2}+\frac{1}{2}x-2x - 1=x^{2}-\frac{3}{2}x - 1\).

Step3: Multiply by third factor

\(P(x)=(x^{2}-\frac{3}{2}x - 1)(x + 3)\)
\[

$$\begin{align*} P(x)&=x^{2}(x + 3)-\frac{3}{2}x(x + 3)-1(x + 3)\\ &=x^{3}+3x^{2}-\frac{3}{2}x^{2}-\frac{9}{2}x-x - 3\\ &=x^{3}+(3x^{2}-\frac{3}{2}x^{2})-(\frac{9}{2}x+x)-3\\ &=x^{3}+\frac{3}{2}x^{2}-\frac{11}{2}x - 3 \end{align*}$$

\]

Step1: Identify roots from graph

The zeros of the graph are \(x=-4,-1,3\). So the polynomial \(Q(x)=k(x + 4)(x + 1)(x - 3)\). Let \(k = 1\).

Step2: Multiply first two factors

\((x + 4)(x + 1)=x^{2}+x+4x + 4=x^{2}+5x + 4\).

Step3: Multiply by third factor

\(Q(x)=(x^{2}+5x + 4)(x - 3)\)
\[

$$\begin{align*} Q(x)&=x^{2}(x - 3)+5x(x - 3)+4(x - 3)\\ &=x^{3}-3x^{2}+5x^{2}-15x+4x - 12\\ &=x^{3}+2x^{2}-11x - 12 \end{align*}$$

\]

Step1: Factor out common factor

Factor out \((x - 2)\) from \(6(x - 2)(x - 3)+4(x - 2)(x - 5)\), we get \((x - 2)[6(x - 3)+4(x - 5)]\).

Step2: Simplify the expression inside the brackets

\[

$$\begin{align*} 6(x - 3)+4(x - 5)&=6x-18 + 4x-20\\ &=(6x + 4x)-(18 + 20)\\ &=10x-38 \end{align*}$$

\]
So the original expression is \((x - 2)(10x - 38)\).

Step3: Substitute \(x=-3\)

When \(x=-3\), \((-3 - 2)[10\times(-3)-38]=(-5)(-30 - 38)=(-5)\times(-68)=340\)

Answer:

The polynomial in factored - form is \(P(x)=(x - 2)(x+\frac{1}{2})(x + 3)\) and in standard form is \(P(x)=x^{3}+\frac{3}{2}x^{2}-\frac{11}{2}x - 3\)

For question 5: