Factorizing 4x4 - 8x2 9: A Comprehensive Guide
When you are faced with a polynomial expression like 4x4 - 8x2 9, determining its factorization can be quite a challenge. In this guide, we cover the step-by-step process, including the use of substitution, the quadratic formula, and the exploration of root factors.
Substitution for Simplification
One effective method to simplify the given expression is by using substitution. By setting y x2, the expression can be rearranged from 4x4 - 8x2 9 to a quadratic form: 4y2 - 8y 9.
Applying the Quadratic Formula
To factorize the quadratic expression 4y2 - 8y 9, we use the quadratic formula:
Quadratic Formula Recap
The quadratic formula for an expression in the form ay2 by c is given by:
y frac{-b pm sqrt{b2 - 4ac}}{2a}
For the expression 4y2 - 8y 9, the coefficients are:
a 4 b -8 c 9The discriminant is calculated as follows:
b2 - 4ac (-8)2 - 4 cdot 4 cdot 9 64 - 144 -80
Since the discriminant -80 is negative, the quadratic equation does not have real roots. This implies that the original expression 4x4 - 8x2 9 cannot be factored over the reals. However, we can still express it in terms of its complex roots using the quadratic formula:
y frac{8 pm sqrt{-80}}{2 cdot 4} frac{8 pm 4isqrt{5}}{8} frac{8}{8} pm ifrac{sqrt{5}}{2} 1 pm ifrac{sqrt{5}}{2}
Thus, y has complex roots, and the quadratic equation can be expressed in its factored form over the complex numbers as:
4(y - 1 ifrac{sqrt{5}}{2})(y - 1 - ifrac{sqrt{5}}{2})
Back Substitution to Obtain the Factorization
After arriving at the factored form over the complex numbers, we substitute y x2 to express the original polynomial in terms of x:
4(x2 - (1 ifrac{sqrt{5}}{2}))(x2 - (1 - ifrac{sqrt{5}}{2}))
Conclusion
In conclusion, the expression 4x4 - 8x2 9 cannot be factored into real linear factors but can be expressed in terms of its complex factors. This process highlights the importance of substitution and the quadratic formula in tackling such problems, as well as the role of complex numbers in the factorization process.