Exercise 58 Page 396

To solve equations with a variable expression raised to a rational exponent, we can raise each side of the equation to the power of the denominator of the rational exponent. Since $\frac{m}{n}\cdot n=m,$ this will eliminate the root. In our case, we will first isolate the variable term and raise each side of the equation to the power of $3.$ We now have a polynomial, and we need to find its roots. The Rational Root Theorem can help us narrow down the possible roots. Let $Q(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0$ be a polynomial with integer coefficients. There are a limited number of possible roots for $Q(x)=0.$

Integer Roots

Integer roots must be factors of $a_0.$ The constant term for our polynomial is $\text{-}352.$ Its factors, and the possible integer roots, are $\pm 1,$ $\pm 2,$ $\pm 4,$ $\pm 8,$ $\pm 11,$ $\pm 16,$ and $\pm 32.$ Let's check them.

$x$	$x^3-22x^2+153x-352$	$P(x)=x^3-22x^2+153x-352$
$1$	$(1{)}^3-22(1{)}^2+153(1)-352$	$\text{-}220\times$
$\text{-}1$	$(\text{-}1{)}^3-22(\text{-}1{)}^2+153(\text{-}1)-352$	$\text{-}528\times$
$2$	$(2{)}^3-22(2{)}^2+153(2)-352$	$\text{-}126\times$
$\text{-}2$	$(\text{-}2{)}^3-22(\text{-}2{)}^2+153(\text{-}2)-352$	$\text{-}754\times$
$4$	$(4{)}^3-22(4{)}^2+153(4)-352$	$\text{-}28\times$
$\text{-}4$	$(\text{-}4{)}^3-22(\text{-}4{)}^2+153(\text{-}4)-352$	$\text{-}1380\times$
$8$	$(8{)}^3-22(8{)}^2+153(8)-352$	$\text{-}24\times$
$\text{-}8$	$(\text{-}8{)}^3-22(\text{-}8{)}^2+153(\text{-}8)-352$	$\text{-}3496\times$
$11$	$(11{)}^3-22(11{)}^2+153(11)-352$	$0✓$
$\text{-}11$	$(\text{-}11{)}^3-22(\text{-}11{)}^2+153(\text{-}11)-352$	$\text{-}6028\times$
$16$	$(16{)}^3-22(16{)}^2+153(16)-352$	$560\times$
$\text{-}16$	$(\text{-}16{)}^3-22(\text{-}16{)}^2+153(\text{-}16)-352$	$\text{-}12528\times$
$32$	$(32{)}^3-22(32{)}^2+153(32)-352$	$14784\times$
$\text{-}32$	$(\text{-}32{)}^3-22(\text{-}32{)}^2+153(\text{-}32)-352$	$\text{-}60544\times$

We can see above that $11$ is a root for the polynomial. Let's now try to find rational roots.

Other Possible Roots

Since $11$ is a solution for the polynomial, $(x-11)$ is factor of the polynomial. Let's use synthetic division to factor out $(x-11).$ Then we can find other roots using the remaining polynomial. Using synthetic division to remove the first root left us with the following polynomial.

Factoring the Remaining Quadratic Factor

We will use the Quadratic Formula to find the remaining factors. To do so, we will need to identify the values of $a,$ $b,$ and $c.$ We can see above that $a=1,$ $b=\text{-}11,$ and $c=32.$ Now, let's substitute these values into the Quadratic Formula to find the remaining roots. The other two roots for the polynomial are $\frac{11+\sqrt{7}i}{2}$ and $\frac{11-\sqrt{7}i}{2}$ which are complex numbers. Therefore, the only real root for the polynomial is $11.$