SOLVING POLYNOMIAL EQUATIONS

Example #1:  Solve for x:  x4– 9x2 + 14 = 0 Step #1:  Let u represent x2and substitute u for x2. (x2)2 – 9(x2) + 14 = 0 express x4 as (x2)2     u2 – 9u + 14 = 0   Step #2:  Factor and solve. (u – 7)(u – 2) = 0   u – 7 = 0          u – 2 = 0       u = 7               u = 2 Step #3:  Remember that u represents x2and replace u with x2. x2 = 7                   x2 = 2   x =                x =

Example #2:  Solve for z:  z5 – 28z3 + 27z = 0 z5 – 28z3 + 27z = 0 factor out a GCF of z     z(z4 – 28z2 + 27) = 0 substitute u for z2     z(u2 – 28u + 27) = 0 factor     z(u – 27)(u – 1) = 0 solve for u     z = 0     u – 27 = 0      u – 1 = 0                            u = 27          u = 1 replace u with z2     z = 0           z2  = 27          z2  = 1 solve for z     z = 0          z =         z = = ± 1   Therefore, the roots of the polynomial z5 – 28z3 + 27z = 0 are 0, 1, –1, , and .

Example #1:  List all the possible rational roots of the function P(x) = 2x3– 11x2 + 12x + 9. According to the Rational Root Theorem, is a root of 2x3– 11x2 + 12x + 9.   If p is a factor of the constant term, 9, and q is a factor of the leading coefficient, 2. Step #1:  Make an organized list of all factors of 9 and 2. factors of  9:  ±1, ±3, ±9   factors of 2:  ±1, ±2   Step #2:  Form all quotients that have factors of 9 in the numerator and factors of 2 in the denominator. These numbers represent all the possible rational zeros of the function.

Name the constant term.  P(x) = 2x³– x² + 2x + 5

List the factors of the constant term.  P(x) = 2x³– x² + 2x + 5

Name the leading coefficient.  P(x) = 2x³– x² + 2x + 5

Stop!  Go to Questions #6-7 about this section, then return to continue on to the next section.


Solving Polynomial Equations Using the Rational Root Theorem

Note:  The following examples are shown using a graphing calculator. The activities can be also be done using an online graphing program.  Click here to navigate to the online grapher.

Example #1:  Find all of the rational roots of 8x3 + 10x2 – 11x + 2 = 0. According to the Rational Root Theorem, is a root of 8x3 + 10x2 – 11x + 2 = 0, if p is a factor of the constant term, 2, and q is a factor of the leading coefficient, 8. Step #1:  Make an organized list of all factors of 2 and 8. factors of  2:  ±1, ±2   factors of 8:  ±1, ±2, ±4, ±8   Step #2:  Form all quotients that have factors of 2 in the numerator and factors of 8 in the denominator. *Some rational numbers are repeated, such as ±, and ±.   Step #3:  Graph the function on your graphing calculator.  Use the button that is marked for the variable x.  Also, to raise a number to the third power, use the button that is located above the key.  The standard viewing screen is the figure on the left.  ZOOMIN is the figure on the right. One root appears to be –2.  Test whether P(–2) = 0 by using synthetic division. Since the remainder is 0, this means that P(–2) = 0 is true and that –2 is a root of the polynomial.   Step #4:  From the graph there appears to be two real zeros between 0 and 1.  Zoom in (  2 ) and use the trace feature to check.  Press and move the cursor around with the arrow keys until the y-value on your screen gets very very close to 0.  , 2: Zoom In, if necessary to locate this point. It looks like there are two real roots, one at and the other at .  Test whether P = 0 and P = 0 by using synthetic division. Since P = 0 and P = 0, and are also roots of the polynomial.  Therefore, there are three roots –2, , and .

For example:  P(x) = (x – 3)(x – 3)(x+1) or P(x) = x3– 5x2 + 3x + 9 touches the x‑axis at (3, 0) but does not cross the x-axis there, as shown in the graphs below.  Therefore, 3 is a double root of P(x) = 0.

Example #2:  Find all of the rational roots of 3x3 – 2x2 – 12x + 8 = 0 According to the Rational Root Theorem, is a root of 3x3 – 2x2 – 12x + 8 = 0, if p is a factor of the constant term, 8, and q is a factor of the leading coefficient, 3. Step #1:  Make an organized list of all factors of 8 and 3. factors of  2:  ±1, ±2, ±4, ±8   factors of 8:  ±1, ±3   Step #2:  Form all quotients that have factors of 8 in the numerator and factors of 3 in the denominator.   Step #3:  Graph the function on your graphing calculator.  Use the button that is marked for the variable x.  Also, to raise a number to the third power, use the button that is located above the key.  The standard viewing screen is the figure on the left.  ZOOMIN is the figure on the right. There appears to be two real roots at –2 and +2.  Test either P(–2) =0 or P(2)=0 by using synthetic division.  We’ll try P(–2)=0. Since the remainder is 0, this means that P(–2) = 0 is true and that –2 is a root of the polynomial. Now we’ll try P(2) = 0. Since the remainder is 0, this means that P(2) = 0 is true and that 2 is a root of the polynomial.   Step #4:  From the graph there appears to be a real zero between 0 and 1.  Zoom in (  2 ) and use the trace feature to check.  Press and move the cursor around with the arrow keys until the y-value on your screen gets very very close to 0.  , 2: Zoom In, if necessary to locate this point. It looks like there is one real root, .  Test whether P = 0 by using synthetic division. Since P = 0, is a root of the polynomial.  Therefore there are 3 roots –2, 2, and .

Example #3:  Find all of the roots of x3 – 4x2 + 6x – 4 = 0. According to the Rational Root Theorem, is a root of x3 – 4x2 + 6x – 4 = 0, if p is a factor of the constant term, –4, and q is a factor of the leading coefficient, 1. Step #1:  Make an organized list of all factors of 8 and 3. factors of  –4:  ±1, ±2, ±4   factors of 1:  ±1   Step #2:  Form all quotients that have factors of 4 in the numerator and factors of 1 in the denominator.   Step #3:  Graph the function on a graphing calculator.  Use the button that is marked for the variable x.  Also, to raise a number to the third power, use the button that is located above the key.  The standard viewing screen is the figure on the left.  ZOOMIN is the figure on the right. The graph crosses the x-axis once.  Therefore there is only one real root.  This root appears to be 2.  Test whether P(2) = 0 by using synthetic division. Since the remainder is 0, this means that P(–2) = 0 is true and that –2 is a root of the polynomial. The resulting numbers 1, –2 and 2 are the coefficients of the quotient.  The quotient will start with an exponent that is one less than the dividend. 1x2 – 2x + 2 This quotient is called a depressed polynomial.  Since the depressed polynomial is quadratic, use the Quadratic Formula to find the other two roots. x2 – 2x + 2               a = 1        b = –2       c =  2 Therefore the roots of the equation are 2, 1 + i and 1 – i.