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# How do you factor ${x^2} - 23x + 132$?

Last updated date: 17th Jul 2024
Total views: 348k
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Answer
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Hint: In this problem, we have to find the factor of the given quadratic equation. we have to factorize the given quadratic equation by splitting the middle term (i.e. $- 23x$) into two parts such that their sum is equal to middle term (i.e. $- 23x$) and product to equal to constant term (i.e. $+ 132$).

Complete step-by-step solution:
This is a problem based on quadratic equations. Quadratic equations are the polynomial equation whose highest degree is $2$. The standard form of any quadratic equation is $a{x^2} + bx + c = 0$, where $a$, $b$ and $c$ are constant real numbers and $x$ is a variable .
If $\alpha$ and $\beta$ are roots of the given quadratic equation. Then $(x - \alpha )$ and $(x - \beta )$ is the factor of the quadratic equation.
i.e. $a{x^2} + bx + c = (x - \alpha )(x - \beta )$.
Now consider the given quadratic equation,
$\Rightarrow {x^2} - 23x + 132$
The middle term of the above quadratic equation is $- 23x$. We break the middle term into two parts ($- 12x$, $- 11x$) such that their sum is $- 23x$ and their product is the constant term i.e. $+ 132$.
$\Rightarrow {x^2} - 12x - 11x + 132$
Taking out $x$ as common in first two terms and $- 11$ as common in last two term we get
$\Rightarrow x(x - 12) - 11(x - 12)$
Taking out $(x - 12)$ as common from both terms we have,
$\Rightarrow (x - 12)(x - 11)$
Hence, ${x^2} - 23x + 132 = (x - 12)(x - 11)$ is the required factorisation of a given quadratic equation.

Note: The standard form of any quadratic equation is $a{x^2} + bx + c = 0$ , where $a$, $b$ and $c$ are constant real numbers and $x$ is a variable .
The quadratic equation can factorise by splitting the middle term.
The product of factors can be written as a quadratic equation.
The roots of the quadratic equation is given by $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.