
How do you factor the expression \[4{{x}^{2}}-8x-5\]?
Answer
548.1k+ views
Hint: In this problem, we have to find the factor of the given expression \[4{{x}^{2}}-8x-5\] by factorization method. We can first split the middle term -8 into two terms. i.e. the x term with its coefficient in such a way that their addition is equal to the middle term i.e. -8x, and multiplication is equal to \[-5\times 4=-10\times 2=-20\], which is the multiplication of first term and the last term. We can then take common terms outside to get the factors.
Complete step by step solution:
We know that the given expression is,
\[4{{x}^{2}}-8x-5\]
We can first split the middle term to form factors.
We have to expand the middle term i.e. the x term with its coefficient in such a way that their addition is equal to the middle term i.e. 16x, and multiplication is equal to 20.
\[\begin{align}
& \Rightarrow -5\times 4=-10\times 2=-20 \\
& \Rightarrow -10+2=-8 \\
\end{align}\]
We can apply the above step
\[\Rightarrow 4{{x}^{2}}-10x+2x-5\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow \left( 4{{x}^{2}}-10x \right)+\left( 2x-5 \right)\]
Now we can take the common terms outside, we get
\[\Rightarrow 2x\left( 2x-5 \right)+1\left( 2x-5 \right)\]
We can again take the common factor first then the remaining terms to make a factor, we get
\[\Rightarrow \left( 2x+1 \right)\left( 2x-5 \right)\]
Therefore, the factors are \[\left( 2x+1 \right)\left( 2x-5 \right)\].
Note: Students make mistakes while taking the common terms outside the equation. We should always remember that when the common terms are multiplied again it should give the previous equation. We should also solve for x to find the factor of the given equation.
Complete step by step solution:
We know that the given expression is,
\[4{{x}^{2}}-8x-5\]
We can first split the middle term to form factors.
We have to expand the middle term i.e. the x term with its coefficient in such a way that their addition is equal to the middle term i.e. 16x, and multiplication is equal to 20.
\[\begin{align}
& \Rightarrow -5\times 4=-10\times 2=-20 \\
& \Rightarrow -10+2=-8 \\
\end{align}\]
We can apply the above step
\[\Rightarrow 4{{x}^{2}}-10x+2x-5\]
We can now take the first two terms and the last two terms to take common terms outside, we get
\[\Rightarrow \left( 4{{x}^{2}}-10x \right)+\left( 2x-5 \right)\]
Now we can take the common terms outside, we get
\[\Rightarrow 2x\left( 2x-5 \right)+1\left( 2x-5 \right)\]
We can again take the common factor first then the remaining terms to make a factor, we get
\[\Rightarrow \left( 2x+1 \right)\left( 2x-5 \right)\]
Therefore, the factors are \[\left( 2x+1 \right)\left( 2x-5 \right)\].
Note: Students make mistakes while taking the common terms outside the equation. We should always remember that when the common terms are multiplied again it should give the previous equation. We should also solve for x to find the factor of the given equation.
Recently Updated Pages
Master Class 8 Social Science: Engaging Questions & Answers for Success

Master Class 8 English: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Master Class 8 Maths: Engaging Questions & Answers for Success

Master Class 8 Science: Engaging Questions & Answers for Success

Master Class 7 English: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest States of India?

What is the theme or message of the poem The road not class 9 english CBSE

What are the major achievements of the UNO class 9 social science CBSE

Explain the importance of pH in everyday life class 9 chemistry CBSE

Differentiate between parenchyma collenchyma and sclerenchyma class 9 biology CBSE

Give 5 examples of refraction of light in daily life

