Question

How do you factor ${{x}^{6}}-124{{x}^{3}}-125$?

Answer 1

The given polynomial ${{x}^{6}}-124{{x}^{3}}-125$ must first be converted to a quadratic polynomial by substituting ${{x}^{3}}=y$. On making this substitution into the given polynomial, we will obtain the polynomial as ${{y}^{2}}-124y-125$. Then this quadratic polynomial can be factored using the middle term splitting method. Then we have to back substitute $y={{x}^{3}}$ to get the polynomial back in terms of the original variable $x$. Then on finally using the algebraic identities ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$, the polynomial will be factored completely.

Complete step by step solution: The polynomial given in the above question is

& \Rightarrow p\left( x \right)={{x}^{6}}-124{{x}^{3}}-125 \\\ & \Rightarrow p\left( x \right)={{\left( {{x}^{3}} \right)}^{2}}-124{{x}^{3}}-125 \\\ \end{aligned}$$ Let us substitute ${{x}^{3}}=y$ in the above polynomial to get $\Rightarrow p\left( y \right)={{y}^{2}}-124y-125$ The product of the first and the third term is equal to $-125{{y}^{2}}$. According to this product, we split the middle term of $-124y$ as $-124y=y-125y$ to get $$\Rightarrow p\left( y \right)={{y}^{2}}+y-125y-125$$ Now, we take $y$ and $-125$ common from the first two and the last two terms respectively to get $$\Rightarrow p\left( y \right)=y\left( y+1 \right)-125\left( y+1 \right)$$ On taking the common factor $$\left( y+1 \right)$$ outside, we get $\Rightarrow p\left( x \right)=\left( y+1 \right)\left( y-125 \right)$ Now, we back substitute $y={{x}^{3}}$ in the above polynomial to get $\Rightarrow p\left( x \right)=\left( {{x}^{3}}+1 \right)\left( {{x}^{3}}-125 \right)$ Now, on putting $1={{1}^{3}}$ and $125={{5}^{3}}$ we get $\Rightarrow p\left( x \right)=\left( {{x}^{3}}+{{1}^{3}} \right)\left( {{x}^{3}}-{{5}^{3}} \right)$ Using the algebraic identities ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ in the first and the second factors respectively of the above polynomial we get $\begin{aligned} & \Rightarrow p\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-x+{{1}^{2}} \right)\left( x-5 \right)\left( {{x}^{2}}+5x+{{5}^{2}} \right) \\\ & \Rightarrow p\left( x \right)=\left( x+1 \right)\left( {{x}^{2}}-x+1 \right)\left( x-5 \right)\left( {{x}^{2}}+5x+25 \right) \\\ & \Rightarrow p\left( x \right)=\left( x+1 \right)\left( x-5 \right)\left( {{x}^{2}}-x+1 \right)\left( {{x}^{2}}+5x+25 \right) \\\ \end{aligned}$ Hence, the given polynomial is completely factored. **Note:** In the above question, we were given a sixth order polynomial to factor. It might be possible to factor it using the hit and trial method, but for a polynomial for such a high degree, it becomes a very lengthy task. So we converted the given polynomial into the quadratic polynomial by substituting ${{x}^{3}}=y$. Since the given polynomial is in terms of $x$ and not in terms of $y$, so do not forget to back substitute $y={{x}^{3}}$.