Solved Examples on the Theorems of Polynomials



Samuel Dominic Chukwuemeka (SamDom For Peace) For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
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For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
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Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.

For Questions (1.) through (10.):
Use your knowledge of Polynomial Theorems to:
(1.) Write the polynomial in factored form.
(2.) Determine the zeros.

Let:
Leading Coefficient = LC
Constant Term = CT
(1.) $f(x) = x^3 - x^2 - 37x - 35$


$ f(x) = x^3 - x^2 - 37x - 35 \\[3ex] Polynomial\;\;is\;\;in\;\;standard\;\;form \\[3ex] \underline{Rational\;\;Zero\;\;Theorem} \\[3ex] LC = 1 \\[3ex] Factors\;\;of\;\;LC = \pm 1 \\[3ex] CT = -35 \\[3ex] Factors\;\;of\;\;CT = \pm 1, \pm 5, \pm 7, \pm 35 \\[3ex] Possible\;\;Rational\;\;Zeros = \dfrac{Factors\;\;of\;\;CT}{Factors\;\;of\;\;LC} \\[5ex] = \dfrac{\pm 1}{\pm 1}, \dfrac{\pm 5}{\pm 1}, \dfrac{\pm 7}{\pm 1}, \dfrac{\pm 35}{\pm 1} \\[5ex] = \pm 1, \pm 5, \pm 7, \pm 35 \\[5ex] Try\;\;x = -1 \\[3ex] f(-1) = (-1)^3 - (-1)^2 - 37(-1) - 35 \\[3ex] f(-1) = -1 - 1 + 37 - 35 \\[3ex] f(-1) = 0 \\[3ex] x = -1 \;\;is\;\;a\;\;zero \\[3ex] x + 1 \;\;is\;\;a\;\;factor \\[3ex] \underline{Synthetic\;\;Division:\;\;By\;\;Addition} \\[3ex] \begin{array}{c|cccc} -1 & 1 & -1 & -37 & -35 \\ + & & -1 & 2 & 35 \\ \hline & 1 & -2 & -35 & 0 \\ \end{array} $
$ Quotient = x^2 - 2x - 35 \\[3ex] = (x + 5)(x - 7) \\[3ex] \implies \\[3ex] (a.)\;\;Factored\;\;Form\;\;of\;\;f(x) = (x + 1)(x + 5)(x - 7) \\[5ex] (b.)\;\; Zeros\;\;of\;\;f(x)\;\;are: \\[3ex] x = -1 \\[3ex] x = -5 \\[3ex] x = 7 $
(2.) $f(x) = 2x^3 - x^2 + 2x - 1$


$ f(x) = 2x^3 - x^2 + 2x - 1 \\[3ex] Polynomial\;\;is\;\;in\;\;standard\;\;form \\[3ex] \underline{Rational\;\;Zero\;\;Theorem} \\[3ex] LC = 2 \\[3ex] Factors\;\;of\;\;LC = \pm 1, \pm 2 \\[3ex] CT = -1 \\[3ex] Factors\;\;of\;\;CT = \pm 1 \\[3ex] Possible\;\;Rational\;\;Zeros = \dfrac{Factors\;\;of\;\;CT}{Factors\;\;of\;\;LC} \\[5ex] = \dfrac{\pm 1}{\pm 1}, \dfrac{\pm 1}{\pm 2} \\[5ex] = \pm 1, \dfrac{\pm 1}{\pm 2} \\[5ex] \underline{Descartes'\;\;Rule\;\;of\;\;Signs} \\[3ex] f(1) = 2(1)^3 - 1^2 + 2(1) - 1 \\[3ex] f(1) = 2(1) - 1 + 2 - 1 \\[3ex] f(1) = 2 \color{darkblue}{-} 1 \color{darkblue}{+} 2 \color{darkblue}{-} 1 \\[3ex] Variations\;\;in\;\;signs = 2\;\;or\;\;0 \\[5ex] f(-1) = 2(-1)^3 - (-1)^2 + 2(-1) - 1 \\[3ex] f(-1) = 2(-1) - 1 - 2 - 1 \\[3ex] f(-1) = -2 - 1 - 2 - 1 \\[3ex] Variations\;\;in\;\;signs = None \\[3ex] $ There is no negative real zero.
So, we shall not waste our time in that regard.

$ Try\;\;x = \dfrac{1}{2} \\[5ex] f\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{2}\right)^3 - \left(\dfrac{1}{2}\right)^2 + 2\left(\dfrac{1}{2}\right) - 1 \\[5ex] f\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{8}\right) - \dfrac{1}{4} + 1 - 1 \\[5ex] f\left(\dfrac{1}{2}\right) = \dfrac{1}{4} - \dfrac{1}{4} + 1 - 1 \\[5ex] f\left(\dfrac{1}{2}\right) = 0 \\[5ex] x = \dfrac{1}{2} \;\;is\;\;a\;\;zero \\[5ex] x - \dfrac{1}{2} \;\;is\;\;a\;\;factor \\[5ex] \underline{Synthetic\;\;Division:\;\;By\;\;Addition} \\[3ex] \begin{array}{c|cccc} \dfrac{1}{2} & 2 & -1 & 2 & -1 \\ + & & 1 & 0 & 1 \\ \hline & 2 & 0 & 2 & 0 \\ \end{array} $
$ Quotient = 2x^2 + 0x + 2 \\[3ex] = 2x^2 + 2 \\[3ex] = 2(x^2 + 1) \\[3ex] $ $x^2 + 1$ cannot be simplified further.

$ \implies \\[3ex] (a.)\;\;Factored\;\;Form\;\;of\;\;f(x) = 2\left(x - \dfrac{1}{2}\right)(x^2 + 1) \\[5ex] (b.)\;\; Zeros\;\;of\;\;f(x)\;\;are: \\[3ex] 1st:\;\;x = \dfrac{1}{2} \\[5ex] x^2 + 1 = 0 \\[3ex] x^2 = -1 \\[3ex] x = \pm \sqrt{-1} \\[3ex] x = \pm i \\[3ex] 2nd:\;\;x = i \\[3ex] 3rd:\;\;x = -i $
(3.)

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(7.)


(8.) Use the Descartes' Rule of Signs and your knowledge of Polynomial Function Theorems to complete the table below.
Write all possibilities.


(9.)

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(11.)

(12.) Use the Descartes' Rule of Signs and your knowledge of Polynomial Function Theorems to complete the table below.
Write all possibilities.

Possible Zeros of Polynomial Functions
Function Number of zeros Possible positive real zeros Possible negative real zeros Complex zeros
... ... ... ... ...


$ (a.)\;\; f(x) = 9x^7 + 2x^6 + x^4 + 4x + 3 \\[3ex] (b.)\;\; f(x) = -3x^3 - 7x^2 + x - 6 \\[3ex] (c.)\;\; f(x) = 5x^4 - 8x^2 - x + 7 \\[3ex] (d.)\;\; f(x) = -6x^7 + x^3 - x^2 + 3 \\[3ex] (e.)\;\; f(x) = -9x^4 + 6x^2 - 3 \\[3ex] $

Possible Zeros of Polynomial Functions
Function Number of zeros Possible positive real zeros Possible negative real zeros Complex zeros
$ (a.)\;\; f(x) = 9x^7 + 2x^6 + x^4 + 4x + 3 \\[3ex] \underline{Positive} \\[3ex] f(1) = 9 + 2 + 1 + 4 + 3 \\[3ex] Variation\;\;in\;\;signs = 0 \\[5ex] \underline{Negative} \\[3ex] f(-1) = 9(-1)^7 + 2(-1)^6 + (-1)^4 + 4(-1) + 3 \\[3ex] f(-1) = -9 + 2 + 1 - 4 + 3 \\[3ex] f(-1) = \color{darkblue}{-}9 \color{darkblue}{+} 2 \color{black}{+} 1 \color{black}{-} 4 \color{black}{+} 3 \\[3ex] Variation\;\;in\;\;signs = 3\;\;or\;\;1 $ 7 0
0 		3
1 		4
6
$ (b.)\;\; f(x) = -3x^3 - 7x^2 + x - 6 \\[3ex] \underline{Positive} \\[3ex] f(1) = -3 \color{darkblue}{-} 7 \color{darkblue}{+} 1 \color{darkblue}{-} 6 \\[3ex] Variation\;\;in\;\;signs = 2\;\;or\;\;0 \\[5ex] \underline{Negative} \\[3ex] f(-1) = -3(-1)^3 - 7(-1)^2 + (-1) - 6 \\[3ex] f(-1) = 3 - 7 - 1 - 6 \\[3ex] f(-1) = \color{darkblue}{+} 3 \color{darkblue}{-} 7 - 1 - 6 \\[3ex] Variation\;\;in\;\;signs = 1 $ 3 2
0 		1
1 		0
2
$ (c.)\;\; f(x) = 5x^4 - 8x^2 - x + 7 \\[3ex] \underline{Positive} \\[3ex] f(1) = 5 - 8 - 1 + 7 \\[3ex] f(1) = \color{darkblue}{+} 5 \color{darkblue}{-} 8 \color{black}{-} 1 \color{black}{+} 7 \\[3ex] Variation\;\;in\;\;signs = 2\;\;or\;\;0 \\[5ex] \underline{Negative} \\[3ex] f(-1) = 5(-1)^4 - 8(-1)^2 - (-1) + 7 \\[3ex] f(-1) = 5 - 8 + 1 + 7 \\[3ex] f(-1) = \color{darkblue}{+} 5 \color{darkblue}{-} 8 \color{darkblue}{+} 1 + 7 \\[3ex] Variation\;\;in\;\;signs = 2\;\;or\;\;0 $ 4 2
2
0
0 		2
0
2
0 		0
2
2
4
$ (d.)\;\; f(x) = -6x^7 + x^3 - x^2 + 3 \\[3ex] \underline{Positive} \\[3ex] f(1) = -6 + 1 - 1 + 3 \\[3ex] f(1) = \color{darkblue}{-} 6 \color{darkblue}{+} 1 \color{darkblue}{-} 1 \color{darkblue}{+} 3 \\[3ex] Variation\;\;in\;\;signs = 3\;\;or\;\;1 \\[5ex] \underline{Negative} \\[3ex] f(-1) = -6(-1)^7 + (-1)^3 - (-1)^2 + 3 \\[3ex] f(-1) = 6 - 1 - 1 + 3 \\[3ex] f(-1) = \color{darkblue}{+} 6 \color{darkblue}{-} 1 \color{black}{-} 1 \color{black}{+} 3 \\[3ex] Variation\;\;in\;\;signs = 2\;\;or\;\;0 $ 7 3
3
1
1 		2
0
2
0 		2
4
4
6
$ (e.)\;\; f(x) = -9x^4 + 6x^2 - 3 \\[3ex] \underline{Positive} \\[3ex] f(1) = -9 + 6 - 3 \\[3ex] f(1) = \color{darkblue}{-} 9 \color{darkblue}{+} 6 \color{darkblue}{-} 3 \\[3ex] Variation\;\;in\;\;signs = 2\;\;or\;\;0 \\[5ex] \underline{Negative} \\[3ex] f(-1) = -9(-1)^4 + 6(-1)^2 - 3 \\[3ex] f(-1) = -9 + 6 - 3 \\[3ex] f(-1) = \color{darkblue}{-} 9 \color{darkblue}{+} 6 \color{darkblue}{-} 3 \\[3ex] Variation\;\;in\;\;signs = 2\;\;or\;\;0 $ 4 2
2
0
0 		2
0
2
0 		0
2
2
4
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