/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 115 Factor completely. $$6 x^{4}+3... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 115

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

Short Answer

Expert verified
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 by step solution

01

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 \(6y^{2}+35y-6=0\).
02

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 \((6y-1)(y+6)=0\)
03

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 \(y+6=0\) => \(y=-6\).
04

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).

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Most popular questions from this chapter

Describe some of the difficulties in factoring polynomials. What suggestions can you offer to overcome these difficulties?

Determine whether each statement 鈥渕akes sense鈥� or 鈥渄oes not make sense鈥� and explain your reasoning. When a factorization requires two factoring techniques, I'm less likely to make errors if I show one technique at a time rather than combining the two factorizations into one step.

Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$72 a^{3} b^{2}+12 a^{2}-24 a^{4} b^{2}$$

Use the \([\mathrm{GRAPH}]\) or \([\mathrm { TABLE }]\) feature of a graphing utility to determine if the polynomial on the left side of each equation has been correctly factored. If not, factor the polynomial correctly and then use your graphing utility to verify the factorization. $$\begin{aligned} &x^{4}-16=\left(x^{2}+4\right)(x+2)(x-2) ;[-5,5,1] \text { by }\\\ &[-20,20,2] \end{aligned}$$

Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$16 x^{4} y-y^{5}$$
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