Finite Math Examples

Identify the Zeros and Their Multiplicities
y=x34x211x+30
Step 1
Set x34x211x+30 equal to 0.
x34x211x+30=0
Step 2
Solve for x.
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Step 2.1
Factor the left side of the equation.
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Step 2.1.1
Factor x34x211x+30 using the rational roots test.
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Step 2.1.1.1
If a polynomial function has integer coefficients, then every rational zero will have the form pq where p is a factor of the constant and q is a factor of the leading coefficient.
p=±1,±30,±2,±15,±3,±10,±5,±6
q=±1
Step 2.1.1.2
Find every combination of ±pq. These are the possible roots of the polynomial function.
±1,±30,±2,±15,±3,±10,±5,±6
Step 2.1.1.3
Substitute 2 and simplify the expression. In this case, the expression is equal to 0 so 2 is a root of the polynomial.
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Step 2.1.1.3.1
Substitute 2 into the polynomial.
23422112+30
Step 2.1.1.3.2
Raise 2 to the power of 3.
8422112+30
Step 2.1.1.3.3
Raise 2 to the power of 2.
844112+30
Step 2.1.1.3.4
Multiply 4 by 4.
816112+30
Step 2.1.1.3.5
Subtract 16 from 8.
8112+30
Step 2.1.1.3.6
Multiply 11 by 2.
822+30
Step 2.1.1.3.7
Subtract 22 from 8.
30+30
Step 2.1.1.3.8
Add 30 and 30.
0
0
Step 2.1.1.4
Since 2 is a known root, divide the polynomial by x2 to find the quotient polynomial. This polynomial can then be used to find the remaining roots.
x34x211x+30x2
Step 2.1.1.5
Divide x34x211x+30 by x2.
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Step 2.1.1.5.1
Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of 0.
x-2x3-4x2-11x+30
Step 2.1.1.5.2
Divide the highest order term in the dividend x3 by the highest order term in divisor x.
x2
x-2x3-4x2-11x+30
Step 2.1.1.5.3
Multiply the new quotient term by the divisor.
x2
x-2x3-4x2-11x+30
+x3-2x2
Step 2.1.1.5.4
The expression needs to be subtracted from the dividend, so change all the signs in x32x2
x2
x-2x3-4x2-11x+30
-x3+2x2
Step 2.1.1.5.5
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2
x-2x3-4x2-11x+30
-x3+2x2
-2x2
Step 2.1.1.5.6
Pull the next terms from the original dividend down into the current dividend.
x2
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
Step 2.1.1.5.7
Divide the highest order term in the dividend 2x2 by the highest order term in divisor x.
x2-2x
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
Step 2.1.1.5.8
Multiply the new quotient term by the divisor.
x2-2x
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
-2x2+4x
Step 2.1.1.5.9
The expression needs to be subtracted from the dividend, so change all the signs in 2x2+4x
x2-2x
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
+2x2-4x
Step 2.1.1.5.10
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2-2x
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
+2x2-4x
-15x
Step 2.1.1.5.11
Pull the next terms from the original dividend down into the current dividend.
x2-2x
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
+2x2-4x
-15x+30
Step 2.1.1.5.12
Divide the highest order term in the dividend 15x by the highest order term in divisor x.
x2-2x-15
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
+2x2-4x
-15x+30
Step 2.1.1.5.13
Multiply the new quotient term by the divisor.
x2-2x-15
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
+2x2-4x
-15x+30
-15x+30
Step 2.1.1.5.14
The expression needs to be subtracted from the dividend, so change all the signs in 15x+30
x2-2x-15
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
+2x2-4x
-15x+30
+15x-30
Step 2.1.1.5.15
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
x2-2x-15
x-2x3-4x2-11x+30
-x3+2x2
-2x2-11x
+2x2-4x
-15x+30
+15x-30
0
Step 2.1.1.5.16
Since the remander is 0, the final answer is the quotient.
x22x15
x22x15
Step 2.1.1.6
Write x34x211x+30 as a set of factors.
(x2)(x22x15)=0
(x2)(x22x15)=0
Step 2.1.2
Factor x22x15 using the AC method.
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Step 2.1.2.1
Factor x22x15 using the AC method.
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Step 2.1.2.1.1
Consider the form x2+bx+c. Find a pair of integers whose product is c and whose sum is b. In this case, whose product is 15 and whose sum is 2.
5,3
Step 2.1.2.1.2
Write the factored form using these integers.
(x2)((x5)(x+3))=0
(x2)((x5)(x+3))=0
Step 2.1.2.2
Remove unnecessary parentheses.
(x2)(x5)(x+3)=0
(x2)(x5)(x+3)=0
(x2)(x5)(x+3)=0
Step 2.2
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
x2=0
x5=0
x+3=0
Step 2.3
Set x2 equal to 0 and solve for x.
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Step 2.3.1
Set x2 equal to 0.
x2=0
Step 2.3.2
Add 2 to both sides of the equation.
x=2
x=2
Step 2.4
Set x5 equal to 0 and solve for x.
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Step 2.4.1
Set x5 equal to 0.
x5=0
Step 2.4.2
Add 5 to both sides of the equation.
x=5
x=5
Step 2.5
Set x+3 equal to 0 and solve for x.
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Step 2.5.1
Set x+3 equal to 0.
x+3=0
Step 2.5.2
Subtract 3 from both sides of the equation.
x=3
x=3
Step 2.6
The final solution is all the values that make (x2)(x5)(x+3)=0 true. The multiplicity of a root is the number of times the root appears.
x=2 (Multiplicity of 1)
x=5 (Multiplicity of 1)
x=3 (Multiplicity of 1)
x=2 (Multiplicity of 1)
x=5 (Multiplicity of 1)
x=3 (Multiplicity of 1)
Step 3
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