8A F11 Q9 - Revision history

Matt Lee at 15:23, 8 April 2015

2015-04-08T15:23:56Z

Matt Lee at 22:33, 6 April 2015

2015-04-06T22:33:33Z

Matt Lee: Created page with "'''Question: ''' a) List all the possible rational zeros of the function $f(x) = x^4 + 5x^3 - 27x^2 +31x -10$
&..."

2015-03-23T18:05:41Z

Created page with "'''Question: ''' a) List all the possible rational zeros of the function <math>f(x) = x^4 + 5x^3 - 27x^2 +31x -10</math><br> &..."

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'''Question: ''' a) List all the possible rational zeros of the function <math>f(x) = x^4 + 5x^3 - 27x^2 +31x -10</math><br>
                 
b) Find all the zeros, that is, solve f(x) = 0

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations
|-
|1) What does the Rational Zeros Theorem say about the possible zeros?
|-
|2) How do you check if a possible zero is actually a zero?
|-
|3) How do you find the rest of the zeros?
|-
|Answer:
|-
|1) The possible divisors can be found by finding the factors of -10, in a list, and the factors of 1, in a second list. Then write down all the fractions with numerators from the first list and denominators from the second list.
|-
|2) Use synthetic division, or plug a possible zero into the function. If you get 0, you have found a zero.
|-
|3) After your reduce the polynomial with synthetic division, try and find another zero from the list you made in part a). Once you reach a degree 2 polynomial you can finish the problem with the quadratic formula.
|}

Solution:

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|Start by factoring -10, and 1. Then the Rational Zeros Theorem gives us that the possible rational zeros are <math>\pm 1, \pm 2, \pm 5,</math> and <math>\pm 10</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
|-
|Start testing zeros with 1 and -1 since they require the least arithmetic. You will also find that 1 is a zero. Applying synthetic division you can reduce the polynomial to <math>x^3 + 6x^2 - 21x + 10</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 3:
|-
|Now we just need to find the zeros of <math>x^3 + 6x^2 - 21x + 10</math>. Since we are not down to a quadratic polynomial we have to continue finding zeros from the list of rational zeros we found in step 1. You will find 2 is another root, and the polynomial can further be reduced to <math>x^2 + 8x - 5</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 4:
|-
|Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are <math>\frac{-8 \pm \sqrt{64 + 20}}{2} = \frac{-8 \pm \sqrt{4*21}}{2} = -4 \pm sqrt{21}</math>. Thus the zeros of <math>x^4 + 5x^3 - 27x^2 + 31x - 10</math> are <math>1, 2, </math>and <math>-4 \pm \sqrt{21}</math>
|}

← Older revision		Revision as of 22:33, 6 April 2015
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	\|Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are <math>\frac{-8 \pm \sqrt{64 + 20}}{2} = \frac{-8 \pm \sqrt{4*21}}{2} = -4 \pm sqrt{21}</math>. Thus the zeros of <math>x^4 + 5x^3 - 27x^2 + 31x - 10</math> are <math>1, 2, </math>and <math>-4 \pm \sqrt{21}</math>		\|Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are <math>\frac{-8 \pm \sqrt{64 + 20}}{2} = \frac{-8 \pm \sqrt{4*21}}{2} = -4 \pm sqrt{21}</math>. Thus the zeros of <math>x^4 + 5x^3 - 27x^2 + 31x - 10</math> are <math>1, 2, </math>and <math>-4 \pm \sqrt{21}</math>
	\|}		\|}
		+
		+	[[8AF11Final\|<u>'''Return to Sample Exam</u>''']]

@@ Line 44: / Line 44: @@
 ! Step 4:
 |-
-|Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are <math>\frac{-8 \pm \sqrt{64 + 20}}{2} = \frac{-8 \pm \sqrt{4*21}}{2} = -4 \pm sqrt{21}</math>. Thus the zeros of <math>x^4 + 5x^3 - 27x^2 + 31x - 10</math> are <math>1, 2, </math>and <math>-4 \pm \sqrt{21}</math>
+|Now that the polynomial has been reduced to a quadratic polynomial we can use the quadratic formula to find the rest of the zeros. By doing so we find the roots are <math>\frac{-8 \pm \sqrt{64 + 20}}{2} = \frac{-8 \pm \sqrt{4*21}}{2} = -4 \pm \sqrt{21}</math>. Thus the zeros of <math>x^4 + 5x^3 - 27x^2 + 31x - 10</math> are <math>1, 2, </math>and <math>-4 \pm \sqrt{21}</math>
 |}
 [[8AF11Final|<u>'''Return to Sample Exam</u>''']]

8A F11 Q9 - Revision history

Matt Lee at 15:23, 8 April 2015

Matt Lee at 22:33, 6 April 2015

Matt Lee: Created page with "'''Question: ''' a) List all the possible rational zeros of the function f(x) = x^4 + 5x^3 - 27x^2 +31x -10 &..."

Matt Lee: Created page with "'''Question: ''' a) List all the possible rational zeros of the function $f(x) = x^4 + 5x^3 - 27x^2 +31x -10$
&..."