How do you find all the zeros of #f(x)=x^4-8x^3+17x^2-8x+16 #?
- If we divide #x^4-8x^3+17x^2-8x+15# by #(x-4)# (using synthetic or long polynomial division) we get factors #(x-4)(x^3-4x^2+x-4)# Again, applying the factor theorem of #x^3-4x^2+x-4)# we find the only zero at #x=4# (again). Dividing #x^3-4x^2+x-4# by #(x-4)# we can re-factor as #(x-4)(x-4)(x^2+1)#.
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1 Answer
Explanation:
2 3 X2 4
Given
#color(white)('XXX')color(red)(1)x^4-8x^3+17x^2-8x+color(blue)(16)#
By the rational root theorem, we know that any rational roots must be factors of #color(blue)(16)/color(red)(1) =16#
which implies they must be in the set#{+-1,+-2,+-4,+-16}#
which implies they must be in the set
Testing each of these possibilities give a zero at (an only at) #x=4#
If we divide #x^4-8x^3+17x^2-8x+15# by #(x-4)# (using synthetic or long polynomial division)
we get factors#(x-4)(x^3-4x^2+x-4)#
we get factors
Again, applying the factor theorem of #x^3-4x^2+x-4)#
we find the only zero at#x=4# (again).
we find the only zero at
Dividing #x^3-4x^2+x-4# by #(x-4)#
we can re-factor as#(x-4)(x-4)(x^2+1)#
we can re-factor as
Related questions
4 3 2 1 Soccer
Step by step solution :
Step 1 :
Equation at the end of step 1 :
Step 2 :
1 X 2
Equation at the end of step 2 :
Step 3 :
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Step 4 :
Equation at the end of step 4 :
Step 5 :
Equation at the end of step 5 :
Step 6 :
Theory - Roots of a product :
6.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
6.2 Solve : 3x = 0
Divide both sides of the equation by 3:
x = 0
Divide both sides of the equation by 3:
x = 0
Solving a Single Variable Equation :
6.3 Solve : x+1 = 0
Subtract 1 from both sides of the equation :
x = -1
Subtract 1 from both sides of the equation :
x = -1
Solving a Single Variable Equation :
6.4 Solve : 2x-7 = 0
Add 7 to both sides of the equation :
2x = 7
Divide both sides of the equation by 2:
x = 7/2 = 3.500
Add 7 to both sides of the equation :
2x = 7
Divide both sides of the equation by 2:
x = 7/2 = 3.500
Solving a Single Variable Equation :
6.5 Solve : (3x+4)2 = 0
(3x+4)2 represents, in effect, a product of 2 terms which is equal to zero
For the product to be zero, at least one of these terms must be zero. Since all these terms are equal to each other, it actually means : 3x+4 = 0
Subtract 4 from both sides of the equation :
3x = -4
Divide both sides of the equation by 3:
x = -4/3 = -1.333
(3x+4)2 represents, in effect, a product of 2 terms which is equal to zero
For the product to be zero, at least one of these terms must be zero. Since all these terms are equal to each other, it actually means : 3x+4 = 0
Subtract 4 from both sides of the equation :
3x = -4
Divide both sides of the equation by 3:
x = -4/3 = -1.333
Solving a Single Variable Equation :
6.6 Solve : x-13 = 0
Add 13 to both sides of the equation :
x = 13
Add 13 to both sides of the equation :
x = 13
Solving a Single Variable Equation :
Butler 4 3 2 X 2.7
6.7 Solve : x+7 = 0
Subtract 7 from both sides of the equation :
x = -7
Subtract 7 from both sides of the equation :
x = -7
6 8 X 1 2 3x 4
6 solutions were found :
Butler 4 3 2 X 2.5
- x = -7
- x = 13
- x = -4/3 = -1.333
- x = 7/2 = 3.500
- x = -1
- x = 0