Question

How do you factor \[{x^2} - 4x - 2\] ?

Answer 1

Hint: The highest exponent of the polynomial in a polynomial equation is called its degree. A polynomial equation has exactly as many roots as its degree. The roots of an equation are the points on the x-axis that is the roots are simply the x-intercepts. In factorization if it’s difficult to split the middle terms we use quadratic formula or Sridhar’s formula that is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] .

Complete step-by-step answer:
The degree of the equation \[{x^2} - 4x - 2 = 0\] is 2, so the number of roots of the given equation is 2.
On comparing the given equation with the standard quadratic equation \[a{x^2} + bx + c = 0\] .
We get \[a = 1\] , \[b = - 4\] and \[c = - 2\] .
For factorization, the standard equation is rewritten as \[a{x^2} + {b_1}x + {b_2}x + c = 0\] .
In the given question, we have to find the value of \[{b_1}\] and \[{b_2}\] by hit and trial method such that \[{b_1} \times {b_2} = - 2\] and \[{b_1} + {b_2} = - 4\] .
But we are not able to find the values of \[{b_1}\] and \[{b_2}\] by hit and trial method, so we find the roots by Quadratic formula.
We know \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Substituting we have,
 \[x = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4(1)( - 2)} }}{{2(1)}}\]
 \[ = \dfrac{{ - 4 \pm \sqrt {16 - 4( - 2)} }}{2}\]
 \[ = \dfrac{{ - 4 \pm \sqrt {16 + 8} }}{2}\]
 \[ = \dfrac{{ - 4 \pm \sqrt {24} }}{2}\]
We can write 24 as \[24 = 6 \times 4\] .
 \[ = \dfrac{{ - 4 \pm \sqrt {4 \times 6} }}{2}\]
We know that square root of 4 is 2 and taking it outside,
 \[ = \dfrac{{ - 4 \pm 2\sqrt 6 }}{2}\]
 \[ = \dfrac{{2( - 2 \pm \sqrt 6 )}}{2}\]
 \[ = - 2 \pm \sqrt 6 \]
Thus we have two roots, they are \[ \Rightarrow x = - 2 + \sqrt 6 \] and \[x = - 2 - \sqrt 6 \],
Hence the factors are \[x - \left( { - 2 + \sqrt 6 } \right)\] and \[x - \left( { - 2 - \sqrt 6 } \right)\] .
So, the correct answer is “ \[x - \left( { - 2 + \sqrt 6 } \right)\] and \[x - \left( { - 2 - \sqrt 6 } \right)\] ”.

Note: In various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis, that is the roots are simply the x-intercepts.