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How do you factor \[{x^2} - 23x + 132\]?

seo-qna
Last updated date: 17th Jul 2024
Total views: 348k
Views today: 5.48k
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 \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].