Question

How do you factor ${x^2} - 5x + 6 = 0$?

Answer 1

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 ${x^2} - 5x + 6 = 0$ 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 $a = 1$, $b = - 5$and $c = 6$.
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 ${b_1} \times {b_2} = 6$ and ${b_1} + {b_2} = - 5$.
That Is we have ${b_1} = - 2$ and ${b_1} = - 3$ because it satisfies the conditions ${b_1} \times {b_2} = 6$ and ${b_1} + {b_2} = - 5$.
Then we have,
$\Rightarrow {x^2} - 2x - 3x + 6 = 0$
Taking ‘x’ common in the first two terms and taking $- 3$from the remaining two terms.
$x(x - 2) - 3(x - 2) = 0$
Again taking $(x - 2)$ common we have,
$\Rightarrow (x - 3)(x - 2) = 0$

Thus the factors are $(x - 3)$ and $(x - 2)$.

Note: We can also find the roots of the above quadratic equation by substituting the obtained factors to zero.
$\Rightarrow (x - 3) = 0$ and $(x - 2) = 0$
$\Rightarrow x = 3$ and $x = 2$ are the roots.
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.