Factor \(x^4-10x^3+25x^2-25\) given there is no value \(x=a\), \(a \in \mathbb{R}\) such that \(P(a)=0\).

Solution

Since there is no \(a\) such that \(P(a)=0\), there can be no linear factors in the factorization of the quartic; we will try to find two quadratic factors.

If we use a modified version of the “have and need” method, we can factor this quartic into two quadratic factors.

The product of the first terms of each quadratic factor must be \(x^4\), so the first term in each factor is \(x^2\).

The product of the last terms of each quadratic factor must be \(-25\). Therefore, the last terms are \(-5\) and \(5\), \(-25\) and \(1\), or \(25\) and \(-1\).

Factor \(x^4-10x^3+25x^2-25\) given there is no value \(x=a\), \(a \in \mathbb{R}\) such that \(P(a)=0\).

Solution

Let's start with \(5\) and \(-5\) and see if we can find a solution.

We can let the coefficients of the middle terms be \(m\) and \(n\).

Expanding the expression gives

Therefore,

Factor \(x^4-10x^3+25x^2-25\) given there is no value \(x=a\), \(a \in \mathbb{R}\) such that \(P(a)=0\).

Solution

Adding \((1)\) \(+\) \((3)\), \(2n=-10\), and therefore \(n=-5\).

Substitution of \(n=-5\) into \((1)\) or \((3)\) results in \(m=-5\).

Having used \((1)\) and \((3)\) to solve the system of equations, it is important to ensure the solution satisfies \((2)\), which it does.

Therefore, \(x^4-10x^3+25x^2-25=(x^2-5x+5)(x^2-5x-5)\).

Factor \(x^4-10x^3+25x^2-25\) given there is no value \(x=a\), \(a \in \mathbb{R}\) such that \(P(a)=0\).

Solution

Another approach is as follows (similar to factoring by grouping).

First, factor \(x^2\) from the first three terms of the quartic, producing a quadratic factor.

We may recognize the three terms of the quadratic as terms of a perfect square with the form similar to

By expressing this quadratic in factored form, we end up with a “difference of perfect squares” for the quartic.

When factoring a polynomial, \(P(x)\), of degree \(3\) or greater, consider the following:

• First, look for a common factor.
• Next, check to see if the polynomial can be factored by grouping. Cubic polynomials, \(ax^3+bx^2+cx+d\), can be factored by grouping if (and only if) \(ad=bc\).
• The factor theorem can be used to determine the first factor of the polynomial. If \(P(a)=0\), then \((x-a)\) is a factor. Long and synthetic division, or the “have and need” method, can be used to find the corresponding factor.
• Polynomials of the form \(ax^{2n}+bx^n+c\) are essentially quadratic in form. Factor these polynomials as you would a quadratic (it may help to make the substitution \(t=x^n\) to obtain \(at^2+bt+c\)).
• To factor the sum or difference of cubes, the following two formulas can be applied:
  \[A^3+B^3=\textcolor{BrickRed}{(A+B)}(A^2-AB+B^2)\]
  \[A^3-B^3=\textcolor{BrickRed}{(A-B)}(A^2+AB+B^2)\]

Note: Not all polynomials can be factored using these methods.