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Factoring trinomials in which the leading term is not 1 is only slightly more difficult than when the leading coefficient is 1. The method used to factor the trinomial is unchanged.
Example 1
Solve for \(x\) in \(x^4 – 13x^2 + 36 = 0\).
First start by converting this trinomial into a form that is more common. Here, it would be a lot easier when factoring \(x^2 – 13x + 36 = 0.\) There is a standard strategy to achieve this through substitution.
First, let \(u = x^2\). Now substitute \(u\) for every \(x^2\), the equation is transformed into \(u^2-13u+36=0\).
\(u^2 – 13u + 36 = 0\) factors into \((u – 9)(u – 4) = 0\).
Once the equation is factored, replace the substitutions with the original variables, which means that, since \(u = x^2\), then \((u – 9)(u – 4) = 0\) becomes \((x^2 – 9)(x^2 – 4) = 0\).
To complete the factorization and find the solutions for \(x\), then \((x^2 – 9)(x^2 – 4) = 0\) must be factored once more. This is done using the difference of squares equation: \(a^2 – b^2 = (a + b)(a – b)\).
Factoring \((x^2 – 9)(x^2 – 4) = 0\) thus leaves \((x – 3)(x + 3)(x – 2)(x + 2) = 0\).
Solving each of these terms yields the solutions \(x = \pm 3, \pm 2\).
This same strategy can be followed to solve similar large-powered trinomials and binomials.
Example 2
Factor the binomial \(x^6 – 7x^3 – 8 = 0\).
Here, it would be a lot easier if the expression for factoring was \(x^2 – 7x – 8 = 0\).
First, let \(u = x^3\), which leaves the factor of \(u^2 – 7u – 8 = 0\).
\(u^2 – 7u – 8 = 0\) easily factors out to \((u – 8)(u + 1) = 0\).
Now that the substituted values are factored out, replace the \(u\) with the original \(x^3\). This turns \((u – 8)(u + 1) = 0\) into \((x^3 – 8)(x^3 + 1) = 0\).
The factored \((x^3 – 8)\) and \((x^3 + 1)\) terms can be recognized as the difference of cubes.
These are factored using \(a^3 – b^3 = (a – b)(a^2 + ab + b^2)\) and \(a^3 + b^3 = (a + b)(a^2 – ab + b^2)\).
And so, \((x^3 – 8)\) factors out to \((x – 2)(x^2 + 2x + 4)\) and \((x^3 + 1)\) factors out to \((x + 1)(x^2 – x + 1)\).
Combining all of these terms yields:
\[(x – 2)(x^2 + 2x + 4)(x + 1)(x^2 – x + 1) = 0\]
The two real solutions are \(x = 2\) and \(x = -1\). Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions.
Questions
Factor each of the following polynomials and solve what you can.
- \(x^4-5x^2+4=0\)
- \(y^4-9y^2+20=0\)
- \(m^4-7m^2-8=0\)
- \(y^4-29y^2+100=0\)
- \(a^4-50a^2+49=0\)
- \(b^4-10b^2+9=0\)
- \(x^4+64=20x^2\)
- \(6z^6-z^3=12\)
- \(z^6-216=19z^3\)
- \(x^6-35x^3+216=0\)
Answers to odd questions
1. \(\text{let }u=x^2 \)
\(\therefore u^2-5u+4=0 \)
\(\text{factors to }(u-4)(u-1)=0 \)
\(\text{replace }u: (x^2-4)(x^2-1)=0 \\ \)
\((x-2)(x+2)(x-1)(x+1)=0 \)
\(x=\pm 2, \pm 1\)
3. \(u=m^2\)
\(\therefore u^2-7u-8=0\)
\((u-8)(u+1)=0\)
\((m^2-8)(m^2+1)=0 \\ \)
\((m+\sqrt{8})(m-\sqrt{8})(m^2+1)=0 \)
\(m=\pm \sqrt{8}\text{ or }\pm 2\sqrt{2}\)
\(m^2+1\text{ has 2 non-real solutions}\)
5. \(\text{let }u=a^2\)
\(\therefore u^2-50u+49=0\)
\((u-49)(u-1)=0\)
\((a^2-49)(a^2-1)=0 \\ \)
\((a-7)(a+7)(a-1)(a+1)=0\)
\(a=\pm 7, \pm1\)
7. \(x^4-20x^2+64=0\)
\(\text{let }u=x^2\)
\(\therefore u^2-20u+64=0\)
\((u-16)(u-4)=0\)
\((x^2-16)(x^2-4)=0 \\ \)
\((x-4)(x+4)(x-2)(x+2)=0\)
\(x=\pm 4, \pm 2\)
9. \(z^6-19z^3-216=0\)
\(\text{let }u=z^3\)
\(\therefore u^2-19u-216=0\)
\((u-27)(u+8)=0\)
\((z^3-27)(z^3+8)=0 \\ \)
\((z-3)(z^2+3z+9)(z+2)(z^2-2z+4)=0\)
\(z=3, -2\)
\(2\text{ non-real solutions each for the 2nd and 4th factors}\)
