How do you factor the trinomial \[{{x}^{2}}-7x+6\]?
Answer
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Hint: Use the middle term split method to factorize \[{{x}^{2}}-7x+6\]. Split -7x into two terms in such a way that their sum is -7x and product is \[6{{x}^{2}}\]. For this process, find the prime factors of 6 and combine them in such a manner so that we can get our condition satisfied. Finally, take the common terms together and write \[{{x}^{2}}-7x+6\] as a product of two terms given as \[\left( x-a \right)\left( x-b \right)\]. Here, ‘a’ and ‘b’ are called zeroes of the polynomial.
Complete step by step answer:
Here, we have been asked to factorize the polynomial: \[{{x}^{2}}-7x+6\] which is quadratic in nature.
Let us use the middle term split method for the factorization.
It says that we have to split the middle term which is -7x into two terms such that their sum is -7x and product is \[6{{x}^{2}}\]. To do this, first we need to find all the prime factors of 6. So, let us find.
We know that 6 can be written as: - \[6=2\times 3\] as the product of its primes. Now, we have to group 2 and 3 such that our conditions of the middle term split method is satisfied. So, we have,
(i) \[\left( -x \right)+\left( -6x \right)=-7x\]
(ii) \[\left( -x \right)\times \left( -6x \right)=6{{x}^{2}}\]
Hence, both the conditions of the middle term split method is satisfied and therefore, the quadratic polynomial can be written as: -
\[\Rightarrow {{x}^{2}}-7x+6={{x}^{2}}-x-6x+6\]
Grouping the term together we have,
\[\Rightarrow {{x}^{2}}-7x+6=x\left( x-1 \right)-6\left( x-1 \right)\]
Taking \[\left( x-1 \right)\] common we have,
\[\Rightarrow {{x}^{2}}-7x+6=\left( x-1 \right)\left( x-6 \right)\]
Hence, \[\left( x-1 \right)\left( x-6 \right)\] is the factored form of the given quadratic polynomial.
Note:
One may note that we can use another method for the factorization. The Discriminant method can also be applied to solve the question. What we will do is we will find the solution of the quadratic equation using discriminant method. The values of x obtained will be assumed as x = a and x = b. Finally, we will consider the product \[\left( x-a \right)\left( x-b \right)\] to get the factored form.
Complete step by step answer:
Here, we have been asked to factorize the polynomial: \[{{x}^{2}}-7x+6\] which is quadratic in nature.
Let us use the middle term split method for the factorization.
It says that we have to split the middle term which is -7x into two terms such that their sum is -7x and product is \[6{{x}^{2}}\]. To do this, first we need to find all the prime factors of 6. So, let us find.
We know that 6 can be written as: - \[6=2\times 3\] as the product of its primes. Now, we have to group 2 and 3 such that our conditions of the middle term split method is satisfied. So, we have,
(i) \[\left( -x \right)+\left( -6x \right)=-7x\]
(ii) \[\left( -x \right)\times \left( -6x \right)=6{{x}^{2}}\]
Hence, both the conditions of the middle term split method is satisfied and therefore, the quadratic polynomial can be written as: -
\[\Rightarrow {{x}^{2}}-7x+6={{x}^{2}}-x-6x+6\]
Grouping the term together we have,
\[\Rightarrow {{x}^{2}}-7x+6=x\left( x-1 \right)-6\left( x-1 \right)\]
Taking \[\left( x-1 \right)\] common we have,
\[\Rightarrow {{x}^{2}}-7x+6=\left( x-1 \right)\left( x-6 \right)\]
Hence, \[\left( x-1 \right)\left( x-6 \right)\] is the factored form of the given quadratic polynomial.
Note:
One may note that we can use another method for the factorization. The Discriminant method can also be applied to solve the question. What we will do is we will find the solution of the quadratic equation using discriminant method. The values of x obtained will be assumed as x = a and x = b. Finally, we will consider the product \[\left( x-a \right)\left( x-b \right)\] to get the factored form.
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