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How do you factor \[{x^2} - 4x - 45\] ?

seo-qna
Last updated date: 21st Jun 2024
Total views: 375k
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Answer
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Hint:This equation is in the form of a quadratic equation \[a{x^2} + bx + c = 0\] . When we solve this equation we either go for a quadratic formula or factorize it and that factorization gives us the roots of the equation. Here when factoring we will check that the factors should be in such a way that their addition is the second term of the equation and their product is the third term of the equation. Then simply we can say that they are the roots of the given equation. So let’s start!

Complete step by step answer:
Given that \[{x^2} - 4x - 45\]
Equating this equation to zero we get,
\[ \Rightarrow {x^2} - 4x - 45 = 0\]
Now the condition of factorization is that the factors on multiplication would give a third term of the equation that is \[ - 45\].
Now factors of \[45\]are \[1 \times 45,3 \times 15,5 \times 9\]
But we now need those two numbers whose difference is 4 only because our second term has coefficient as \[ - 4\].
So we will choose \[5 \times 9\] . But the signs should be matching with our requirements. So we need one of them to be negative as the product is \[ - 45\] . Then we take \[ - 9\& 5\] . Now let’s check,
\[ \Rightarrow {x^2} - 4x - 45 = 0\]
On factoring,
\[ \Rightarrow {x^2} - 9x + 5x - 45 = 0\]
Taking x common from first two terms and 5 common from last two terms,
\[ \Rightarrow x\left( {x - 9} \right) + 5\left( {x - 9} \right) = 0\]
Taking in brackets,
\[ \Rightarrow \left( {x - 9} \right)\left( {x + 5} \right) = 0\]
Equating these brackets to zero we get,
\[ \Rightarrow \left( {x - 9} \right) = 0,\left( {x + 5} \right) = 0\]
On solving this
\[ \Rightarrow x = 9,x = - 5\]
These are our factors.

Note: Students in order to cross check out any of these values in the equation again.
\[ \Rightarrow {x^2} - 4x - 45 = 0\]
Let’s put \[x = - 5\]
\[ \Rightarrow {\left( { - 5} \right)^2} - 4\left( { - 5} \right) - 45\]
On solving,
\[ \Rightarrow 25 + 20 - 45\]
\[ \Rightarrow 45 - 45 = 0\]
Same number subtracted from the same number gives zero and that is our RHS. So yes the roots are correct. In these types of problems be careful about the signs and brackets. As in addition as multiplications of two signs it matters so much. Note the table given below.

SignsAdditionMultiplication
\[ + , + \]\[ + \]\[ + \]
\[ + , - \]\[ + \] ((first number is greater)\[ - \]
\[ - , + \]\[ - \] (negative(negative number is greater)\[ - \]
\[ - , - \]\[ - \]\[ + \]



Note that the numbers we select for factoring have exactly opposite signs as that of the factors so obtained.