To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[/latex].

[latex]\begin{array}{l}\\ 2\overline{)\begin{array}{l}6\hfill & -1\hfill & -15\hfill & 2\hfill & -7\hfill \\ \hfill & 12\hfill & \text{ }22\hfill & 14\hfill & 32\hfill \end{array}}\\ \begin{array}{l}\text{ }6\hfill & 11\hfill & \text{ }7\hfill & \text{16}\hfill & 25\hfill \end{array}\end{array}[/latex]

The remainder is 25. Therefore, [latex]f\left(2\right)=25[/latex].

Analysis of the Solution

We can check our answer by evaluating [latex]f\left(2\right)[/latex].

[latex]\begin{array}{l}f\left(x\right) & =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\ f\left(2\right) & =6{\left(2\right)}^{4}-{\left(2\right)}^{3}-15{\left(2\right)}^{2}+2\left(2\right)-7 \\ \hfill & =25\hfill \end{array}[/latex]

The only possible rational zeros of [latex]f\left(x\right)[/latex] are the quotients of the factors of the last term, –4, and the factors of the leading coefficient, 2.

The constant term is –4; the factors of –4 are [latex]p=\pm 1,\pm 2,\pm 4[/latex].

The leading coefficient is 2; the factors of 2 are [latex]q=\pm 1,\pm 2[/latex].

If any of the four real zeros are rational zeros, then they will be of one of the following factors of –4 divided by one of the factors of 2.

[latex]\begin{array}{l}\frac{p}{q}=\pm \frac{1}{1},\pm \frac{1}{2}\text{ }& \frac{p}{q}=\pm \frac{2}{1},\pm \frac{2}{2}\text{ }& \frac{p}{q}=\pm \frac{4}{1},\pm \frac{4}{2}\end{array}[/latex]

Note that [latex]\frac{2}{2}=1[/latex] and [latex]\frac{4}{2}=2[/latex], which have already been listed. So we can shorten our list.

[latex]\frac{p}{q}=\frac{\text{Factors of the last}}{\text{Factors of the first}}=\pm 1,\pm 2,\pm 4,\pm \frac{1}{2}[/latex]

The Rational Zero Theorem tells us that if [latex]\frac{p}{q}[/latex] is a zero of [latex]f\left(x\right)[/latex], then p is a factor of 1 and q is a factor of 2.

[latex]\begin{array}{l}\frac{p}{q}=\frac{\text{factor of constant term}}{\text{factor of leading coefficient}}\hfill \\ \text{ }=\frac{\text{factor of 1}}{\text{factor of 2}}\hfill \end{array}[/latex]

The factors of 1 are [latex]\pm 1[/latex] and the factors of 2 are [latex]\pm 1[/latex] and [latex]\pm 2[/latex]. The possible values for [latex]\frac{p}{q}[/latex] are [latex]\pm 1[/latex] and [latex]\pm \frac{1}{2}[/latex]. These are the possible rational zeros for the function. We can determine which of the possible zeros are actual zeros by substituting these values for x in [latex]f\left(x\right)[/latex].

[latex]\begin{array}{l}\text{ }f\left(-1\right)=2{\left(-1\right)}^{3}+{\left(-1\right)}^{2}-4\left(-1\right)+1=4\hfill \\ \text{ }f\left(1\right)=2{\left(1\right)}^{3}+{\left(1\right)}^{2}-4\left(1\right)+1=0\hfill \\ \text{ }f\left(-\frac{1}{2}\right)=2{\left(-\frac{1}{2}\right)}^{3}+{\left(-\frac{1}{2}\right)}^{2}-4\left(-\frac{1}{2}\right)+1=3\hfill \\ \text{ }f\left(\frac{1}{2}\right)=2{\left(\frac{1}{2}\right)}^{3}+{\left(\frac{1}{2}\right)}^{2}-4\left(\frac{1}{2}\right)+1=-\frac{1}{2}\hfill \end{array}[/latex]

Of those, [latex]-1,-\frac{1}{2},\text{ and }\frac{1}{2}[/latex] are not zeros of [latex]f\left(x\right)[/latex]. 1 is the only rational zero of [latex]f\left(x\right)[/latex].

We can use synthetic division to show that [latex]\left(x+2\right)[/latex] is a factor of the polynomial.

The remainder is zero, so [latex]\left(x+2\right)[/latex] is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:

[latex]\left(x+2\right)\left({x}^{2}-8x+15\right)[/latex]

We can factor the quadratic factor to write the polynomial as

[latex]\left(x+2\right)\left(x - 3\right)\left(x - 5\right)[/latex]

By the Factor Theorem, the zeros of [latex]{x}^{3}-6{x}^{2}-x+30[/latex] are –2, 3, and 5.