Question

How do you factor \[{{x}^{2}}-x-6\] ?

Answer 1

Hint: Let consider a quadratic equation \[a{{x}^{2}}-bx+c\] , during factorisation of such quadratic equation we use completing the perfect square method that always work, in this method we make this this equation a complete square of a linear equation by just doing some modifications in it.
First equate the above quadratic polynomial with zero, this will lead to an equation and when we put a value in variable \[x\] then it satisfies the equation and makes both sides zero.
Now divide both sides by \[a\] then the coefficient of highest degree will be one the just making it a perfect square this will leads to \[{{\left( x+\dfrac{b}{2a} \right)}^{2}}-{{\left( \dfrac{b}{2a} \right)}^{2}}+\dfrac{{{c}^{2}}}{{{a}^{2}}}=0\].

Complete step-by-step answer:
Since in this quadratic polynomial \[{{x}^{2}}-x-6\], the coefficient of highest degree is \[1\]
Now equating this quadratic polynomial with zero
\[\Rightarrow {{x}^{2}}-x-6=0\]
Then we don’t need to divide it with its coefficient, as it will be the same on dividing with \[1\]
Now to make its perfect square of a linear equation
Now add \[\dfrac{1}{4}\] to both sides
\[\Rightarrow {{x}^{2}}-x-6+\dfrac{1}{4}=0+\dfrac{1}{4}\]
\[\Rightarrow {{x}^{2}}-x+\dfrac{1}{4}-6=\dfrac{1}{4}\]
\[\Rightarrow {{(x-\dfrac{1}{2})}^{2}}=6+\dfrac{1}{4}\]
\[\Rightarrow {{(x-\dfrac{1}{2})}^{2}}=\dfrac{25}{4}\]
\[\Rightarrow (x-\dfrac{1}{2})=\pm \sqrt{\dfrac{25}{4}}\]
\[\Rightarrow x=\dfrac{1}{2}\pm \dfrac{5}{2}=\dfrac{1\pm 5}{2}\]
Taking positive sign,
\[x=3\]
Taking negative sign, \[x=-2\]
Hence the two roots of the given quadratic equation are \[-2,3\]
And we know that when the roots of a quadratic equation are \[a,b\] then the equation is written as
\[(x-a)(x-b)=0\] where \[(x-a)\] and \[(x-b)\] are the factors of quadratic equation
Hence the factors of given quadratic equation are \[(x+2)\] and \[(x-3)\].

Note: When the coefficient of highest degree is \[1\] then guess the two numbers whose product is \[c\] and the sum will be \[-b\] in standard form of equation where \[a=1\] Then these two numbers will be the roots or zeros of equation. It is suggested to use the completing the perfect square method as it is quite simple.