Certainly! Let's factorize the polynomial [tex]\( h(x) = 2x^4 + 7x^3 - 17x^2 - 7x + 15 \)[/tex] completely step by step.
### Step 1: Identify Possible Rational Roots
To factorize the polynomial completely, we start by checking for possible rational roots. These roots are potential candidates derived from the Rational Root Theorem, which states that any rational root, p/q, of a polynomial with integer coefficients is such that [tex]\( p \)[/tex] is a factor of the constant term and [tex]\( q \)[/tex] is a factor of the leading coefficient.
For the polynomial [tex]\( h(x) = 2x^4 + 7x^3 - 17x^2 - 7x + 15 \)[/tex]:
- The constant term (last term) is 15. The factors of 15 are [tex]\( \pm 1, \pm 3, \pm 5, \pm 15 \)[/tex].
- The leading coefficient (first term) is 2. The factors of 2 are [tex]\( \pm 1, \pm 2 \)[/tex].
So, the possible rational roots are [tex]\( \pm 1, \pm \frac{1}{2}, \pm 3, \pm \frac{3}{2}, \pm 5, \pm \frac{5}{2}, \pm 15, \pm \frac{15}{2} \)[/tex].
### Step 2: Testing Possible Rational Roots
We substitute each possible rational root into the polynomial [tex]\( h(x) = 2x^4 + 7x^3 - 17x^2 - 7x + 15 \)[/tex] to see if it equals zero.
After testing, we find that the roots [tex]\( x = 1 \)[/tex], [tex]\( x = -1 \)[/tex], [tex]\( x = -5 \)[/tex], and [tex]\( x = \frac{3}{2} \)[/tex] satisfy the polynomial equation.
### Step 3: Factoring Using Known Roots
Knowing these roots, we can factorize the polynomial as follows:
- Since [tex]\( x = 1 \)[/tex] is a root, [tex]\( (x - 1) \)[/tex] is a factor.
- Since [tex]\( x = -1 \)[/tex] is a root, [tex]\( (x + 1) \)[/tex] is a factor.
- Since [tex]\( x = -5 \)[/tex] is a root, [tex]\( (x + 5) \)[/tex] is a factor.
- Since [tex]\( x = \frac{3}{2} \)[/tex] is a root, [tex]\( (2x - 3) \)[/tex] is a factor.
Thus, the original polynomial can be factored completely as:
[tex]\[ h(x) = (x - 1)(x + 1)(x + 5)(2x - 3) \][/tex]
### Conclusion
The completely factorized form of the polynomial [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]
