4. Write the equation of a polynomial of degree 3 (in factored form if possible), whose graph passes through the points $A(4,-5), B(-1,0), C(0,5)$ and $D(5,0)$. Sketch the graph of the polynomial.
4. Points $B$ and $D$ are zeros of the polynomial, so we can write $y=a(x+1)(x-5)\left(x-x_{3}\right)$. The polynomial intersects the y-axis at point $C(0,5)$, so $5=a \cdot 1 \cdot(-5) \cdot\left(-x_{3}\right)$ or $5 a x_{3}=5$, from which we get $x_{3}=\frac{1}{a}$, since $a \neq 0$. Finally, considering that the polynomial passes through point $A$, we have $-5=a \cdot 5 \cdot(-1) \cdot\left(4-\frac{1}{a}\right)$ or $a \cdot\left(4-\frac{1}{a}\right)=1$, from which we express $a=\frac{1}{2}$. The polynomial has the equation $y=$ $\frac{1}{2}(x+1)(x-5)(x-2)$.

Notation: $y=a\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)$

or $y=a(x+1)(x-5)\left(x-x_{3}\right) \ldots \ldots \ldots .1$ point

Point $C$ - written equation $a \cdot x_{3}=1 \ldots \ldots \ldots . .1$ point

Point $A$ - written equation $1=4 a-a x_{3} \ldots \ldots .1$ point

Solutions $a=\frac{1}{2}, x=2 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \ldots$ point

Polynomial notation: $y=\frac{1}{2}(x+1)(x-5)(x-2) \ldots .1$ point

Graph sketch

1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_06886e410b39c0216094g-11.jpg?height=942&width=622&top_left_y=1125&top_left_x=1225)