
Factorize: $ {y^2} - 5y + 6 $
Answer
588.9k+ views
Hint: For the equation of the form $ a{x^2} + bx + c = 0 $ , if $ m,n $ are the roots of the equation , then \[m + n = \dfrac{{ - b}}{a}\] and $ m \cdot n = \dfrac{c}{a} $ . Use the substitution method to solve the equations \[m + n = \dfrac{{ - b}}{a}\] and $ m \cdot n = \dfrac{c}{a} $ . After obtaining the values of $ m,{\rm{ }}n $ write it in the form of $ \left( {x - m} \right)\left( {x - n} \right) $ which would be the factorised form of the expression $ a{x^2} + bx + c $ .
Complete step-by-step answer:
First check whether the coefficient of $ {y^2} $ is 1.
The expression $ {y^2} - 5y + 6 $ has 1 as the coefficient of $ {y^2} $ .
This expression is to be written in the form of $ \left( {y - m} \right)\left( {y - n} \right) $ , where $ m.{\rm{ }}n $ are the factors of the equation $ {y^2} - \left( {m + n} \right)y + mn $ .
Now, we should compare $ {y^2} - 5y + 6 $ with $ {y^2} - \left( {m + n} \right)y + mn $ .
On comparing, we get, $ m + n = 5 $ and $ mn = 6 $
If $ m = 2 $ and $ n = 3 $ , then $ 2 + 3 = 5,{\rm{ }}2 \times 3 = 6 $ .
Hence, the expression $ {y^2} - 5y + 6 = \left( {y - m} \right)\left( {y - n} \right) $
Here, substitute the values of $ m = 2 $ and $ n = 3 $ in the equation
$\Rightarrow {y^2} - 5y + 6 = \left( {y - m} \right)\left( {y - n} \right) $
Therefore, $ {y^2} - 5y + 6 $ can be factorised as $ {y^2} - 5y + 6 = \left( {y - 2} \right)\left( {y - 3} \right) $ .
Additional Information:
Factorisation of polynomials is done by separating it in the product of factors which cannot be reduced further. It can also be said as the process of converting addition and subtraction of terms into the form of the product of the terms. Factorisation is necessary to convert the complex looking expressions into the product of linear terms.
In 1793, Theodor von Schubert produced the first algorithm for the factorization of the polynomial. The linear factors are being deduced from the polynomials in order to factorise the polynomials.
Note: Students often find it difficult to find the factors of the equation. So, it is important to keep in mind that for the equation of the form $ {x^2} - mx + n $ , if the roots are $ a,{\rm{ }}b $ then $ a + b = m $ and $ a \cdot b = n $ .
We need to just compare the given expression to $ {x^2} - mx + n $ and convert it in the form of $ \left( {x - a} \right)\left( {x - b} \right) $ .
Complete step-by-step answer:
First check whether the coefficient of $ {y^2} $ is 1.
The expression $ {y^2} - 5y + 6 $ has 1 as the coefficient of $ {y^2} $ .
This expression is to be written in the form of $ \left( {y - m} \right)\left( {y - n} \right) $ , where $ m.{\rm{ }}n $ are the factors of the equation $ {y^2} - \left( {m + n} \right)y + mn $ .
Now, we should compare $ {y^2} - 5y + 6 $ with $ {y^2} - \left( {m + n} \right)y + mn $ .
On comparing, we get, $ m + n = 5 $ and $ mn = 6 $
If $ m = 2 $ and $ n = 3 $ , then $ 2 + 3 = 5,{\rm{ }}2 \times 3 = 6 $ .
Hence, the expression $ {y^2} - 5y + 6 = \left( {y - m} \right)\left( {y - n} \right) $
Here, substitute the values of $ m = 2 $ and $ n = 3 $ in the equation
$\Rightarrow {y^2} - 5y + 6 = \left( {y - m} \right)\left( {y - n} \right) $
Therefore, $ {y^2} - 5y + 6 $ can be factorised as $ {y^2} - 5y + 6 = \left( {y - 2} \right)\left( {y - 3} \right) $ .
Additional Information:
Factorisation of polynomials is done by separating it in the product of factors which cannot be reduced further. It can also be said as the process of converting addition and subtraction of terms into the form of the product of the terms. Factorisation is necessary to convert the complex looking expressions into the product of linear terms.
In 1793, Theodor von Schubert produced the first algorithm for the factorization of the polynomial. The linear factors are being deduced from the polynomials in order to factorise the polynomials.
Note: Students often find it difficult to find the factors of the equation. So, it is important to keep in mind that for the equation of the form $ {x^2} - mx + n $ , if the roots are $ a,{\rm{ }}b $ then $ a + b = m $ and $ a \cdot b = n $ .
We need to just compare the given expression to $ {x^2} - mx + n $ and convert it in the form of $ \left( {x - a} \right)\left( {x - b} \right) $ .
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

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

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Trending doubts
What are gulf countries and why they are called Gulf class 8 social science CBSE

Name the states through which the Tropic of Cancer class 8 social science CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

What are the 12 elements of nature class 8 chemistry CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Who created the image of Bharat Mata for the first class 8 social science CBSE

