1. Example-1
1. `x^2+3x-4`, find Rational Zeros Theorem to find all possible rational roots
Solution:
The Given Polynomial `=x^2+3x-4`
`=x^2+3x-4`
If a polynomial has integer coefficients, then every rational zero will have the form `p/q`, where `p` is a factor of the constant and `q` is a factor of the leading coefficient.
Step-1: Find the constant and its factors (both positive and negative)
The constant term is 4
and its factor are, `p=+-1,+-2,+-4`
Step-2: Find the leading coefficient and its factors (both positive and negative)
The leading coefficient is 1
and its factor are, `q = +-1`
Step-3: Find all possible values of `+-p/q`
When `q=+-1` then `p/q=+-1/1,+-2/1,+-4/1`
Step-4: Simplify and remove any duplicates
The possible rational roots are `+-1,+-2,+-4`
Step-5: Now substitute these possible roots in f(x) and if `f(x)=0` then it is the actual rational roots (actual rational zeros)
`f(x)=x^2+3x-4`
1. `f(1)=(1)^2+3(1)-4=0`
Hence, `1` is a root.
2. `f(-1)=(-1)^2+3(-1)-4=-6`
3. `f(2)=(2)^2+3(2)-4=6`
4. `f(-2)=(-2)^2+3(-2)-4=-6`
5. `f(4)=(4)^2+3(4)-4=24`
6. `f(-4)=(-4)^2+3(-4)-4=0`
Hence, `-4` is a root.
The actual rational roots are `1,-4`
2. `3x^2+6x-1`, find Rational Zeros Theorem to find all possible rational roots
Solution:
The Given Polynomial `=3x^2+6x-1`
`=3x^2+6x-1`
If a polynomial has integer coefficients, then every rational zero will have the form `p/q`, where `p` is a factor of the constant and `q` is a factor of the leading coefficient.
Step-1: Find the constant and its factors (both positive and negative)
The constant term is 1
and its factor are, `p=+-1`
Step-2: Find the leading coefficient and its factors (both positive and negative)
The leading coefficient is 3
and its factor are, `q = +-1,+-3`
Step-3: Find all possible values of `+-p/q`
When `q=+-1` then `p/q=+-1/1`
When `q=+-3` then `p/q=+-1/3`
Step-4: Simplify and remove any duplicates
The possible rational roots are `+-1,+-1/3`
Step-5: Now substitute these possible roots in f(x) and if `f(x)=0` then it is the actual rational roots (actual rational zeros)
`f(x)=3x^2+6x-1`
1. `f(1)=3(1)^2+6(1)-1=8`
2. `f(-1)=3(-1)^2+6(-1)-1=-4`
3. `f(1/3)=3(1/3)^2+6(1/3)-1=4/3`
4. `f(-1/3)=3(-1/3)^2+6(-1/3)-1=-8/3`
The actual rational roots are ``
3. `10x^2+6x-1`, find Rational Zeros Theorem to find all possible rational roots
Solution:
The Given Polynomial `=10x^2+6x-1`
`=10x^2+6x-1`
If a polynomial has integer coefficients, then every rational zero will have the form `p/q`, where `p` is a factor of the constant and `q` is a factor of the leading coefficient.
Step-1: Find the constant and its factors (both positive and negative)
The constant term is 1
and its factor are, `p=+-1`
Step-2: Find the leading coefficient and its factors (both positive and negative)
The leading coefficient is 10
and its factor are, `q = +-1,+-2,+-5,+-10`
Step-3: Find all possible values of `+-p/q`
When `q=+-1` then `p/q=+-1/1`
When `q=+-2` then `p/q=+-1/2`
When `q=+-5` then `p/q=+-1/5`
When `q=+-10` then `p/q=+-1/10`
Step-4: Simplify and remove any duplicates
The possible rational roots are `+-1,+-1/2,+-1/5,+-1/10`
Step-5: Now substitute these possible roots in f(x) and if `f(x)=0` then it is the actual rational roots (actual rational zeros)
`f(x)=10x^2+6x-1`
1. `f(1)=10(1)^2+6(1)-1=15`
2. `f(-1)=10(-1)^2+6(-1)-1=3`
3. `f(1/2)=10(1/2)^2+6(1/2)-1=9/2`
4. `f(-1/2)=10(-1/2)^2+6(-1/2)-1=-3/2`
5. `f(1/5)=10(1/5)^2+6(1/5)-1=3/5`
6. `f(-1/5)=10(-1/5)^2+6(-1/5)-1=-9/5`
7. `f(1/10)=10(1/10)^2+6(1/10)-1=-3/10`
8. `f(-1/10)=10(-1/10)^2+6(-1/10)-1=-3/2`
The actual rational roots are ``
4. `2x^2-3x+1`, find Rational Zeros Theorem to find all possible rational roots
Solution:
The Given Polynomial `=2x^2-3x+1`
`=2x^2-3x+1`
If a polynomial has integer coefficients, then every rational zero will have the form `p/q`, where `p` is a factor of the constant and `q` is a factor of the leading coefficient.
Step-1: Find the constant and its factors (both positive and negative)
The constant term is 1
and its factor are, `p=+-1`
Step-2: Find the leading coefficient and its factors (both positive and negative)
The leading coefficient is 2
and its factor are, `q = +-1,+-2`
Step-3: Find all possible values of `+-p/q`
When `q=+-1` then `p/q=+-1/1`
When `q=+-2` then `p/q=+-1/2`
Step-4: Simplify and remove any duplicates
The possible rational roots are `+-1,+-1/2`
Step-5: Now substitute these possible roots in f(x) and if `f(x)=0` then it is the actual rational roots (actual rational zeros)
`f(x)=2x^2-3x+1`
1. `f(1)=2(1)^2-3(1)+1=0`
Hence, `1` is a root.
2. `f(-1)=2(-1)^2-3(-1)+1=6`
3. `f(1/2)=2(1/2)^2-3(1/2)+1=0`
Hence, `1/2` is a root.
4. `f(-1/2)=2(-1/2)^2-3(-1/2)+1=3`
The actual rational roots are `1,1/2`
5. `x^3-2x^2-x+2`, find Rational Zeros Theorem to find all possible rational roots
Solution:
The Given Polynomial `=x^3-2x^2-x+2`
`=x^3-2x^2-x+2`
If a polynomial has integer coefficients, then every rational zero will have the form `p/q`, where `p` is a factor of the constant and `q` is a factor of the leading coefficient.
Step-1: Find the constant and its factors (both positive and negative)
The constant term is 2
and its factor are, `p=+-1,+-2`
Step-2: Find the leading coefficient and its factors (both positive and negative)
The leading coefficient is 1
and its factor are, `q = +-1`
Step-3: Find all possible values of `+-p/q`
When `q=+-1` then `p/q=+-1/1,+-2/1`
Step-4: Simplify and remove any duplicates
The possible rational roots are `+-1,+-2`
Step-5: Now substitute these possible roots in f(x) and if `f(x)=0` then it is the actual rational roots (actual rational zeros)
`f(x)=x^3-2x^2-x+2`
1. `f(1)=(1)^3-2(1)^2-(1)+2=0`
Hence, `1` is a root.
2. `f(-1)=(-1)^3-2(-1)^2-(-1)+2=0`
Hence, `-1` is a root.
3. `f(2)=(2)^3-2(2)^2-(2)+2=0`
Hence, `2` is a root.
4. `f(-2)=(-2)^3-2(-2)^2-(-2)+2=-12`
The actual rational roots are `1,-1,2`
This material is intended as a summary. Use your textbook for detail explanation.
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