How do you factor ${{x}^{2}}+5x-36$ ?
Answer
573k+ views
Hint: From the question given, we have been asked to factor the quadratic expression ${{x}^{2}}+5x-36$. We can find the factors for the given quadratic expression by using the process of factorization. By using the process of factorization, we can find the factors for the given quadratic expression. First of all, we have to know about the process of factorization.
Complete step-by-step solution:
By using the product of coefficients and writing it as the sum of the two numbers we will find out the factors. This is the process of factorization.
From the question given, we have been given a quadratic expression ${{x}^{2}}+5x-36$
First of all, we have to multiply the coefficient of \[{{x}^{2}}\] and constant.
By multiplying the coefficient of \[{{x}^{2}}\] and constant, we get \[\Rightarrow 1\times -36=-36\]
Now, we have to write the coefficient of \[x\] as the sum of two numbers and the numbers should be the factors of \[36\].
\[\Rightarrow -36=9\times -4\]
By writing the given expression as above said, we get \[\Rightarrow {{x}^{2}}+9x-4x-36\]
Now, we have to take the common terms out and simplify further to obtain the factors.
By taking the common terms out and simplifying further, we get
\[\Rightarrow x\left( x+9 \right)-4\left( x+9 \right)\]
\[\Rightarrow \left( x-4 \right)\left( x+9 \right)\]
Therefore, \[{{x}^{2}}+5x-36=\left( x-4 \right)\left( x+9 \right)\]
Hence, we got the factors for the given quadratic expression by using the process of factorization.
Note: We should be well aware of the process of factorization. Also, we should be very careful while writing the middle term as a sum of two numbers which is equal to the product of coefficient of first term and constant. Also, we should be very careful while simplifying the expression. This question can also be answered by using the formulae for finding the roots of any quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . For ${{x}^{2}}+5x-36$ it is given as
$\begin{align}
& \dfrac{-\left( 5 \right)\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 1 \right)\left( -36 \right)}}{2\left( 1 \right)}=\dfrac{5\pm \sqrt{{{5}^{2}}+144}}{2} \\
& \Rightarrow \dfrac{-5\pm \sqrt{25+144}}{2}=\dfrac{-5\pm \sqrt{169}}{2} \\
& \Rightarrow \dfrac{-5\pm 13}{2}= -9,4 \\
\end{align}$
As the roots are $9,-4$ the factors will be $\left( x-4 \right)$ and $\left( x+9 \right)$.
Complete step-by-step solution:
By using the product of coefficients and writing it as the sum of the two numbers we will find out the factors. This is the process of factorization.
From the question given, we have been given a quadratic expression ${{x}^{2}}+5x-36$
First of all, we have to multiply the coefficient of \[{{x}^{2}}\] and constant.
By multiplying the coefficient of \[{{x}^{2}}\] and constant, we get \[\Rightarrow 1\times -36=-36\]
Now, we have to write the coefficient of \[x\] as the sum of two numbers and the numbers should be the factors of \[36\].
\[\Rightarrow -36=9\times -4\]
By writing the given expression as above said, we get \[\Rightarrow {{x}^{2}}+9x-4x-36\]
Now, we have to take the common terms out and simplify further to obtain the factors.
By taking the common terms out and simplifying further, we get
\[\Rightarrow x\left( x+9 \right)-4\left( x+9 \right)\]
\[\Rightarrow \left( x-4 \right)\left( x+9 \right)\]
Therefore, \[{{x}^{2}}+5x-36=\left( x-4 \right)\left( x+9 \right)\]
Hence, we got the factors for the given quadratic expression by using the process of factorization.
Note: We should be well aware of the process of factorization. Also, we should be very careful while writing the middle term as a sum of two numbers which is equal to the product of coefficient of first term and constant. Also, we should be very careful while simplifying the expression. This question can also be answered by using the formulae for finding the roots of any quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ is given by $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ . For ${{x}^{2}}+5x-36$ it is given as
$\begin{align}
& \dfrac{-\left( 5 \right)\pm \sqrt{{{\left( 5 \right)}^{2}}-4\left( 1 \right)\left( -36 \right)}}{2\left( 1 \right)}=\dfrac{5\pm \sqrt{{{5}^{2}}+144}}{2} \\
& \Rightarrow \dfrac{-5\pm \sqrt{25+144}}{2}=\dfrac{-5\pm \sqrt{169}}{2} \\
& \Rightarrow \dfrac{-5\pm 13}{2}= -9,4 \\
\end{align}$
As the roots are $9,-4$ the factors will be $\left( x-4 \right)$ and $\left( x+9 \right)$.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

