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I want to factor this polynomial without using the quadratic formula. So, as the first step I multiply coefficient of $x^2$ with $-10$. The product obtained is $-50$. However, there are no two factors of $-50$ that add up to $-2$ and give the product as $-50$. So, it becomes evident that the polynomial would be difficult to factor by splitting the middle term. So, according to my level of understanding, the next method I would try is by expressing the polynomial as the difference of two squares and use the identity $a^2 - b^2 = (a + b) (a - b)$. Here is my first try at expressing it as the difference of two squares:

$$5x^2 - 2x - 10$$ $$= 4x^2 - 4x + 1 + x^2 + 2x - 11 $$ $$= (2x - 1)^2 + x^2 + 2x - 11$$

I am not able to solve beyond this. So, I tried the same method but with different manipulation of the polynomial:

$$5x^2 - 2x - 10$$ $$= 9x^2 - 6x + 1 - 4x^2 + 4x - 11$$ $$= (3x - 1)^2 - 4x^2 + 4x - 11$$

Again, I can't go beyond this. So, I want to know whether I am using the wrong method or am I just not able to decide the next manipulation of the polynomial?

Rodrigo de Azevedo
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Parth
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5 Answers5

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$5x^2-2x-10=5(x^2-\frac25x-2)=5\left((x-\frac15)^2-\frac{51}{25}\right)= 5\left((x-\frac15+\frac{\sqrt{51}}5)(x-\frac15-\frac{\sqrt{51}}5)\right)$.

AdLibitum
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$$5x^2 - 2x - 10=0$$ $$5(x^2-\frac{2}{5}x-2)=0$$ $$5(x^2-\frac{2}{5}x-2+\frac{1}{25}-\frac{1}{25})=0$$ $$5(x^2-\frac{2}{5}x+\frac{1}{25})-5*2-5*\frac{1}{25}=0$$ $$5(x-\frac{1}{5})^2-\frac{51}{5}=0$$ $$(x-\frac{1}{5})^2-\frac{51}{25}=0$$ $$(x-\frac{1}{5})^2-(\frac{\sqrt51}{5})^2=0$$ now use the fact of $$(a^2-b^2)=(a-b)(a+b)$$ so $$(x-\frac{1}{5}-\frac{\sqrt51}{5})(x-\frac{1}{5}+\frac{\sqrt51}{5})=0$$

E.H.E
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$$5x^2-2x-10=5\left(x^2-\frac25x-2\right)=5\left(x^2-\frac25x+\frac1{25}-2-\frac1{25}\right)=5\left((x-\frac15)^2-\frac{51}{25}\right)=$$ $$=5\left(\left(x-\frac15\right)^2-\left(\frac{\sqrt{51}}{5}\right)^2\right)$$

Roman83
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I want to know whether I am using the wrong method or am I just not able to decide the next manipulation of the polynomial?

$(1)$ $$\begin{align}5x^2 - 2x - 10\\ = 4x^2 - 4x + 1 + x^2 + 2x - 11 \\ = (2x - 1)^2 + x^2 + 2x - 11 \end{align}$$

$(2)$ $$\begin{align}5x^2 - 2x - 10\\ = 9x^2 - 6x + 1 - 4x^2 + 4x - 11\\ = (3x - 1)^2 - 4x^2 + 4x - 11 \end{align}$$

To be able to factorise $(1)$,

By Factor Theorem,

For $$p_1(x)=x^2+2x-11$$ We must have $$p_1\left(\frac{1}{2}\right)=0$$

Clearly, $$p_1\left(\frac{1}{2}\right) \neq 0$$

So, this method yields nothing useful.

In fact, for $$g_1(x)=5x^2-2x-10$$ We observe that $$g_1\left(\frac{1}{2}\right) \neq 0$$

So, $k\left(x-\frac{1}{2}\right)$ or specifically $2x-1$ can't be a factor of $g_1(x)$

Note: A similar argument follows for $(2)$, which is left as an exercise to the reader.

Factor theorem provides the motivation for checking simple points and then hoping that if we get one root we get the other easily.

Let $f(x)=5x^2-2x-10$

Using Rational Root Thorem, we get that any rational root of the form $\dfrac{p}{q}$ can have $p \in \{\pm 1, \pm 2, \pm 5, \pm 10\}$ and $q \in \{\pm 1, \pm 2, \pm 5\}$

After checking through all the cases, we get $f(x)$ doesn't have any rational roots.

Factorisation of quadratics in general

Let $$5x^2-2x-10=(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$$ where $a,b,c,d \in \mathbb{R}$

Comparing like terms we get

$$ac=5\tag{1}$$

$$ad+bc=-2\tag{2}$$

$$bd=-10\tag{3}$$

Note: Although there appear less equations than variables. It may be noted that it is enough to find the ratio among $a,b,c,d$. For example, $$x^2-3x-2=(x-1)(x-2)=(2x-2)\left(\frac{x}{2}-1\right)$$

So, set $a=1$ and solve the $3$ equations to get the value of the remaining $3$ variables.

$\color{red}{\text{Warning: }}$ Since solving linear equations in many variable can be quite tedious, it is best if factorisation is done, in a similar manner, as answered (earlier) by other users.

Roby5
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$$5x^2 - 2x - 10 = \begin{bmatrix} 1\\ x\end{bmatrix}^T \begin{bmatrix} -10 & -1+t\\ -1-t & 5\end{bmatrix} \begin{bmatrix} 1\\ x\end{bmatrix}$$

We want to find a $t$ that makes the matrix above rank-$1$. Computing the determinant,

$$\det \begin{bmatrix} -10 & -1+t\\ -1-t & 5\end{bmatrix} = t^2 - 51$$

If $t = \sqrt{51}$, then the matrix is singular and, thus, rank-$1$. Computing the eigendecomposition,

$$\begin{bmatrix} -10 & -1+\sqrt{51}\\ -1-\sqrt{51} & 5\end{bmatrix} = -5 \begin{bmatrix} \frac{\sqrt{51}-1}{5}\\ 1\end{bmatrix} \begin{bmatrix} \frac{\sqrt{51}+1}{5}\\ -1\end{bmatrix}^T = 5 \begin{bmatrix} \frac{\sqrt{51}-1}{5}\\ 1\end{bmatrix} \begin{bmatrix} -\frac{\sqrt{51}+1}{5}\\ 1\end{bmatrix}^T$$

Thus,

$$\begin{array}{rl} 5x^2 - 2x - 10 &= \begin{bmatrix} 1\\ x\end{bmatrix}^T \begin{bmatrix} -10 & -1+\sqrt{51}\\ -1-\sqrt{51} & 5\end{bmatrix} \begin{bmatrix} 1\\ x\end{bmatrix}\\ &= 5 \begin{bmatrix} 1\\ x\end{bmatrix}^T \begin{bmatrix} \frac{\sqrt{51}-1}{5}\\ 1\end{bmatrix} \begin{bmatrix} -\frac{\sqrt{51}+1}{5}\\ 1\end{bmatrix}^T\begin{bmatrix} 1\\ x\end{bmatrix}\\ &= 5 \left( x - \dfrac{1}{5} + \dfrac{\sqrt{51}}{5} \right) \left( x - \dfrac{1}{5} - \dfrac{\sqrt{51}}{5} \right)\\ &= \dfrac{1}{5} \left( 5 x - 1 + \sqrt{51} \right) \left( 5 x - 1 - \sqrt{51} \right)\end{array}$$

Rodrigo de Azevedo
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