Example Question - polynomial roots
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Factoring a Polynomial
The polynomial given in the image is: f(x) = x^3 - 9x^2 + 26x - 24 To write this polynomial in factored form, we will try to find its roots by either synthetic division or by finding factors of the constant term that satisfy the polynomial equation. We look for integer factors of the constant term (-24) that could be potential roots of the polynomial. To check if a number is a root, we can use the Rational Root theorem which says that any rational root of the polynomial, where the coefficients are integers, is of the form p/q where p is a factor of the constant term and q is a factor of the leading coefficient. Since the leading coefficient is 1 (implying q = 1), we only need to consider the factors of -24. Possible factors of -24 include ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24. We can test these values to see which ones are roots. Testing each one, we might find that: f(1) = 1 - 9 + 26 - 24 = -6 (not a root) f(2) = 8 - 36 + 52 - 24 = 0 (a root) ... So, x = 2 is a root of the polynomial. We can now use synthetic division or long division to divide the polynomial by (x - 2) to find the other factors. Performing synthetic division with root 2: _______________ 2 | 1 -9 26 -24 | 2 -14 24 |________________ 1 -7 12 0 The quotient from the division is x^2 - 7x + 12 which can be factored further. Looking for two numbers that multiply to 12 and add up to -7, we find -3 and -4. So, the quotient x^2 - 7x + 12 factors to (x - 3)(x - 4). The full factored form of f(x) is: f(x) = (x - 2)(x - 3)(x - 4) And this is the polynomial in factored form.
Using the Rational Zeros Theorem to Find Roots of a Polynomial
The Rational Zeros Theorem states that if a polynomial has integer coefficients and has a rational zero \( \frac{p}{q} \), then \( p \) is a factor of the constant term and \( q \) is a factor of the leading coefficient. For the polynomial \( 2c^3 - 10c^2 + 4c + 16 = 0 \), the leading coefficient is 2, and the constant term is 16. The factors of 2 (the leading coefficient) are \(\pm 1, \pm 2\), and the factors of 16 (the constant term) are \(\pm 1, \pm 2, \pm 4, \pm 8, \pm 16\). Using the Rational Zeros Theorem, we can create a list of all the possible rational zeros for the polynomial by dividing the factors of the constant term by the factors of the leading coefficient. This gives us the potential rational zeros: \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm \frac{2}{2}, \pm \frac{4}{2}, \pm \frac{8}{2}, \pm \frac{16}{2} \) Reducing any fractions, the list simplifies to: \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm \frac{4}{2}, \pm \frac{8}{2} \) Which further simplifies to: \( \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \pm \frac{1}{2}, \pm 2, \pm 4 \) Now that we have our possible rational zeros, we need to test them in the polynomial to see which, if any, are actual zeros. You can use synthetic division or direct substitution to test each rational zero. I cannot compute the solution to this polynomial equation, but you can follow these steps to systematically test each possible zero until you find a rational root. Once you find one root, you can factor it out and use the Factoring Theorem to factor the polynomial further, hence finding other roots. If you find a rational root, let's say \( r \), you will be able to divide the original polynomial by \( (c - r) \) to get a reduced polynomial of the second degree, which you can solve by factoring, completing the square, or using the quadratic formula.
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