Sure, let's factorize the polynomial [tex]\( h(x) = 2x^4 + 7x^3 - 17x^2 - 7x + 15 \)[/tex] completely.
First, we should attempt to find the roots of the polynomial. If we can identify the roots, we can express [tex]\( h(x) \)[/tex] as a product of its linear factors.
1. Check for Rational Roots:
We can apply the Rational Root Theorem which suggests that any rational root of the polynomial, [tex]\( p/q \)[/tex], is a factor of the constant term (15) divided by a factor of the leading coefficient (2).
Factors of the constant term [tex]\( 15 \)[/tex]: [tex]\(\pm 1, \pm 3, \pm 5, \pm 15\)[/tex]
Factors of the leading coefficient [tex]\( 2 \)[/tex]: [tex]\(\pm 1, \pm 2\)[/tex]
Potential rational roots could be:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}, \pm 5, \pm \frac{5}{2}, \pm 15 \][/tex]
2. Test Rational Roots by Substitution:
After testing these potential roots, we find that the polynomial [tex]\( h(x) \)[/tex] has the following roots: [tex]\( x = 1, x = -1, x = -5, x = \frac{3}{2} \)[/tex].
3. Construct Factors from Roots:
Each root [tex]\( r \)[/tex] gives us a corresponding factor of [tex]\( x - r \)[/tex]. Since we have the roots [tex]\( 1, -1, -5, \frac{3}{2} \)[/tex]:
- For [tex]\( x = 1 \)[/tex], we get factor [tex]\( x - 1 \)[/tex],
- For [tex]\( x = -1 \)[/tex], we get factor [tex]\( x + 1 \)[/tex],
- For [tex]\( x = -5 \)[/tex], we get factor [tex]\( x + 5 \)[/tex],
- For [tex]\( x = \frac{3}{2} \)[/tex], we get factor [tex]\( 2x - 3 \)[/tex] (to eliminate the fraction).
Hence, the factors of the polynomial [tex]\( h(x) \)[/tex] are:
[tex]\[ h(x) = (x - 1)(x + 1)(x + 5)(2x - 3) \][/tex]
Thus, the completely factorized form of [tex]\( h(x) = 2x^4 + 7x^3 - 17x^2 - 7x + 15 \)[/tex] is:
[tex]\[ h(x) = (x - 1)(x + 1)(x + 5)(2x - 3) \][/tex]
