Question

if f(x) is a polynomial with rational coefficients and is of degree 2 or 3 ,then show that f(x) is reducible if and only if f(x)=0 has a rational root.​

Answer 1

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Answer 2

Answer:

The polynomial is given by $f(x) = (x - 2)(x^2 + 1)(x^2 + 9)$

Step-by-step explanation:

Given: if f(x) is a polynomial with rational coefficients and is of degree 2 or 3.

To find: Show that f(x) is reducible if and only if f(x)=0 has a rational root.​

Solution:

Complex roots come in conjugates.  Your roots are

$x = -3i\\x= 3i\\x = 2\\x = -i\\x = i$

Since$i=\sqrt{-1}$, then $i^{2}=-1$

For the root   $x = 3i$ , we can say that the factor is  

$(x - 3i)(x + 3i) = x^2 - 9i^2\\ = x^2 - 9(-1)\\ = x^2 + 9$

$(x - i)(x + i)\\ = x^2 - i^2\\ = x^2 + 1$

Then The polynomial is$f(x) = (x - 2)(x^2 + 1)(x^2 + 9)$

Method 2:

• If the coefficients are real, and some zeros are complex, then the complex conjugate of each complex zero must also be a zero.  Therefore, $-3i\ and\ i$are also zeros.
• That makes 5 zeros, so the minimum degree is 5.  With a leading coefficient of 1, we get

$f(x) = (x - 3i)(x + 3i)(x - 2)(x + i)(x - i)$

Note that

$(x - 3i)(x + 3i) = x^2 + 9$

$(x + i)(x - i) = x^2 + 1$

Then

$f(x) = (x^2 + 9)(x - 2)(x^2 + 1)$

Note the coefficients are all real, in spite of the 4 complex zeros.