Problem 115 - Introductory Algebra for College Students

Question

Factor completely. $$6 x^{4}+35 x^{2}-6$$

Step-by-Step Solution

Answer

The complete factorization of the equation \(6x^{4}+35x^{2}-6\) is \((\sqrt{6}x-1)(\sqrt{6}x+1)(x+ \sqrt{6}i)(x- \sqrt{6}i)\).

Step 1: Recognize the Quadratic Form

Recognize that the equation \(6x^{4}+35x^{2}-6\) can be viewed as a quadratic equation in the form of \(ax^{2} + bx + c = 0\), where \(x^{2}\) is considered as a single variable (say 'y'). So, the given equation becomes \(6y^{2}+35y-6=0\).

Step 2: Factorize the Quadratic Equation

Factorize the quadratic equation \(6y^{2}+35y-6=0\) by the method of splitting the middle term. To do this, find two numbers that multiply to \(6* -6 = -36\) (i.e., product of 'a' and 'c') and add to 35 (i.e., 'b'). These numbers are 36 and -1. So, we can write 35y as \((36y - y)\). The equation is then written as \(6y^{2}+36y-y-6=0\), which can be further factored to \(6y(y+6)-1(y+6)=0\). This gives us \((6y-1)(y+6)=0\)

Step 3: Find the Roots of the Equation

Solve for y to get the roots of the equation. We get \(6y-1=0\) => \(y=\frac{1}{6}\) and \(y+6=0\) => \(y=-6\).

Step 4: Substitute Back y as \(x^{2}\)

Remember 'y' was a substitute for \(x^{2}\). So, before finding the final answer, substitute back \(y\) as \(x^{2}\) and solve the equations: \(x^{2}=\frac{1}{6}\) (which gives \(x=\sqrt{\frac{1}{6}}\) and \(x=-\sqrt{\frac{1}{6}}\)) and \(x^{2}= -6\) (which gives no real roots, as square roots of negative numbers are imaginary in the real number system).