Answer:
[tex]\textsf{Factors:}\quad (x-1),\;(x-3),\;\textsf{and}\;(-4x-1)[/tex]
[tex]\textsf{Zeros:}\quad 1,\;3,\;\textsf{and}\;-\frac{1}{4}[/tex]
[tex]\textsf{$x$-intercepts:}\quad (1, 0), \;(3, 0),\;\textsf{and} \;\left(-\frac{1}{4},0)[/tex]
Step-by-step explanation:
Given polynomial:
[tex]f(x)=-4x^3+15x^2-8x-3[/tex]
To use synthetic division to test for rational zeros of the polynomial f(x), we need to try dividing the polynomial by potential rational roots using synthetic division.
The Rational Root Theorem states that any rational root (zero) of a polynomial with integer coefficients is a quotient of a factor of the constant term divided by a factor of the leading coefficient.
In this case:
- The constant term is -3, and its factors are ±1 and ±3.
- The leading coefficient is -4, and its factors are ±1, ±2 and ±4.
Therefore, the potential rational zeros are:
[tex]\sf \pm 1, \pm \dfrac{1}{2}, \pm \dfrac{1}{4}, \pm 3, \pm \dfrac{3}{2}, \pm \dfrac{3}{4}[/tex]
Now, perform synthetic division for each potential rational zero.
First, let's try x = 1:
[tex]\begin{array}{c|rrrr}1&-4&15&-8&-3\\\cline{1-1}&\downarrow&-4&11&3\\\cline{2-5}&-4&11&3&0\end{array}[/tex]
The remainder is zero, so x = 1 is a zero of f(x).
Now, perform synthetic division on the quotient -4x² + 11x + 3.
Let's try x = 3:
[tex]\begin{array}{c|rrr}3&-4&11&3\\\cline{1-1}&\downarrow&-12&-3\\\cline{2-4}&-4&-1&0\end{array}[/tex]
The remainder is zero, so x = 3 is also a zero of f(x).
The quotient of the second synthetic division (-4x - 1) is linear, so we can stop here. Therefore, the factored form of the given polynomial is:
[tex]f(x)=(x-1)(x-3)(-4x-1)[/tex]
To find the zeros, set each factor equal to zero and solve for x:
[tex]x-1=0 \implies x=1[/tex]
[tex]x-3=0 \implies x=3[/tex]
[tex]-4x-1=0 \implies x=-\dfrac{1}{4}[/tex]
The x-intercepts, also known as zeros or roots, of a function are the points at which the graph intersects the x-axis, meaning the y-coordinate is zero. Therefore, the x-intercepts are:
[tex](1, 0), \;(3, 0), \;\left(-\frac{1}{4},0)[/tex]
