How do you find the factor of the expression \[{{x}^{2}}-5x-24\] ?
Answer
627.3k+ views
Hint: For the given question we are given to factorize the equation \[{{x}^{2}}-5x-24\] . First of all we have to find the factors of constant and we have to check which pair is suitable for conditions and then we have to split the term -5x in the basis of the selected pair and then solve the equation to get the factors.
Complete step by step solution:
For the given problem we are given to find the factor of the given equation \[{{x}^{2}}-5x-24\] .
To find the factors of the given equation we have to assume the given equation is equal to 0.
Let us consider the given equation as equation (1).
\[{{x}^{2}}-5x-24=0..........\left( 1 \right)\]
Now we have to factorize the equation (1).
In order to factor we have to find factors of the constant and we have to observe which of those add to equal the coefficient before the x value.
By observing the factors of 24 we can clearly say that 8 and 3 are suitable for above conditions.
Therefore, let us rewrite the term \[-5x\] as \[-8x+3x\] .
By rewriting the equation, we get
\[\Rightarrow {{x}^{2}}-8x+3x-24=0\]
Let us consider the given equation as equation (2).
\[{{x}^{2}}-8x+3x-24=0..............\left( 2 \right)\]
Pulling out x from first two terms and 3 from next two terms we get
\[\begin{align}
& \Rightarrow x\left( x-8 \right)+3\left( x-8 \right)=0 \\
& \Rightarrow \left( x+3 \right)\left( x-8 \right)=0 \\
\end{align}\]
Let us consider
\[\left( x+3 \right)\left( x-8 \right)=0..........\left( 3 \right)\]
Therefore \[\left( x+3 \right)\left( x-8 \right)=0\] is the factor of the equation \[{{x}^{2}}-5x-24\] .
Note: Students have to check the conditions while splitting the \[bx\] term (\[a{{x}^{2}}+bx+c=0\]). Examiners may ask roots for quadratic equations, so students have to practise all the types of problems in quadratic equation concept. Students can also do this problem with a direct division method.
Complete step by step solution:
For the given problem we are given to find the factor of the given equation \[{{x}^{2}}-5x-24\] .
To find the factors of the given equation we have to assume the given equation is equal to 0.
Let us consider the given equation as equation (1).
\[{{x}^{2}}-5x-24=0..........\left( 1 \right)\]
Now we have to factorize the equation (1).
In order to factor we have to find factors of the constant and we have to observe which of those add to equal the coefficient before the x value.
By observing the factors of 24 we can clearly say that 8 and 3 are suitable for above conditions.
Therefore, let us rewrite the term \[-5x\] as \[-8x+3x\] .
By rewriting the equation, we get
\[\Rightarrow {{x}^{2}}-8x+3x-24=0\]
Let us consider the given equation as equation (2).
\[{{x}^{2}}-8x+3x-24=0..............\left( 2 \right)\]
Pulling out x from first two terms and 3 from next two terms we get
\[\begin{align}
& \Rightarrow x\left( x-8 \right)+3\left( x-8 \right)=0 \\
& \Rightarrow \left( x+3 \right)\left( x-8 \right)=0 \\
\end{align}\]
Let us consider
\[\left( x+3 \right)\left( x-8 \right)=0..........\left( 3 \right)\]
Therefore \[\left( x+3 \right)\left( x-8 \right)=0\] is the factor of the equation \[{{x}^{2}}-5x-24\] .
Note: Students have to check the conditions while splitting the \[bx\] term (\[a{{x}^{2}}+bx+c=0\]). Examiners may ask roots for quadratic equations, so students have to practise all the types of problems in quadratic equation concept. Students can also do this problem with a direct division method.
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