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If there are any integer roots, they must be factors of the constant term. If there are any rational roots, they have the form ± pq, where p is a factor of the constant term and q a factor of the leading coefficient.
1, - 2, - 2/3
One way of finding the roots for P(x)=0 is to guess and check. This is inefficient unless there is a way to minimize the number of possible roots. The Rational Root Theorem helps us with this!
Let Q(x)= a_nx^n+a_(n-1)x^(n-1)+... +a_1x+ a_0 be a polynomial with integer coefficients. There are a limited number of possible roots for Q(x)=0.
Now let's consider the given polynomial. P(x)= 3x^4+2x^3-9x^2 + 4 We can check for integer and rational roots one at a time.
The constant term of this polynomial is 4. Its factors, and the possible integer roots for P(x)=0, are ± 1, ± 2, and ± 4. Let's check them! Remember, a root will give us a result of 0 after substituted into the polynomial.
| x | 3x^4+2x^3-9x^2+4 | P(x)=3x^4+2x^3-9x^2+4 |
|---|---|---|
| 1 | 3( 1)^4+2( 1)^3-9( 1)^2+4 | 0 âś“ |
| - 1 | 3( - 1)^4+2( - 1)^3-9( - 1)^2+4 | - 4 * |
| 2 | 3( 2)^4+2( 2)^3-9( 2)^2+4 | 32 * |
| - 2 | 3( - 2)^4+2( - 2)^3-9( - 2)^2+4 | 0 âś“ |
| 4 | 3( 4)^4+2( 4)^3-9( 4)^2+4 | 756 * |
| - 4 | 3( - 4)^4+2( - 4)^3-9( - 4)^2+4 | 500 * |
Looking at the table, we can see that 1 and - 2 are roots for P(x)=0.
Next, let's try to find non-integer rational roots. The leading coefficient is 3 and the constant term is 4. Therefore, the possible rational roots are ± 1 3, ± 2 3, and ± 4 3.
| x | 3x^4+2x^3-9x^2+4 | P(x)=3x^4+2x^3-9x^2+4 |
|---|---|---|
| 1/3 | 3( 1/3 )^4+2( 1/3 )^3-9( 1/3 )^2+4 | 3.11 * |
| - 1/3 | 3(- 1/3 )^4+2(- 1/3 )^3-9(- 1/3 )^2+4 | 2.96 * |
| 2/3 | 3( 2/3 )^4+2( 2/3 )^3-9( 2/3 )^2+4 | 1.19 * |
| - 2/3 | 3(- 2/3 )^4+2(- 2/3 )^3-9(- 2/3 )^2+4 | 0 âś“ |
| 4/3 | 3( 4/3 )^4+2( 4/3 )^3-9( 4/3 )^2+4 | 2.22 * |
| - 4/3 | 3(- 4/3 )^4+2(- 4/3 )^3-9(- 4/3 )^2+4 | - 7.26 * |
We found one rational root for P(x)=0, x=- 23.
The rational roots of P(x)=0 include 1, - 2, and - 23.