
Factorize the given polynomial
$6{x^2} - 5x - 6$
Answer
579.3k+ views
Hint: In this particular question use the concept of factorization, the factorization is done on the basis of coefficient of x, so we have to factorize the multiplication of constant term and coefficient of ${x^2}$ such that its sum gives us the coefficient of x and product gives us the multiplication of coefficient of ${x^2}$ and the constant term so use these concepts to reach the solution of the question.
Complete step by step answer:
Given equation
$6{x^2} - 5x - 6$
As we see that the highest power of x is 2 so it is a quadratic equation, if the highest power was 3 then it is a cubic equation.
Now factorize the given quadratic equation, so factorize the multiplication of constant term and coefficient of ${x^2}$ such that the sum or difference gives us -5 and product gives us $\left( {6 \times - 6} \right) = - 36$
As the coefficient of ${x^2}$ is 6, coefficient of x is negative 5 and constant term is -6, so factors of multiplication of constant term and coefficient of ${x^2}$ i.e. $\left( {6 \times - 6} \right) = - 36$, such that the sum of the factors will give us the coefficient of x i.e. -5 and the product of the factors will give us $\left( {6 \times - 6} \right) = - 36$
So -36 can be factorize as -9 and 4,
So the sum of -9 and 4 is -5.
And the product of -9 and 4 is -36.
So the given quadratic equation is written as
$ \Rightarrow 6{x^2} - 9x + 4x - 6$
Now take 3x common from first two terms and 2 from last two terms we have,
$ \Rightarrow 3x\left( {2x - 3} \right) + 2\left( {2x - 3} \right)$
Now take (2x – 3) common we have,
$ \Rightarrow \left( {2x - 3} \right)\left( {3x + 2} \right)$
So this is the required factorization of the given quadratic equation.
Note:
We can also solve this quadratic equation directly by using the quadratic formula which is given as, $\left( {x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right)$, where a, b and c are the coefficients of ${x^2}$, x and constant term respectively, so on comparing, a = 6, b = -5 and c= -6, so simply substitute these values in the given equation and simplify we will get the values of x let the values of x be p and q, so the factors are (x – p) and (x – q) respectively.
Complete step by step answer:
Given equation
$6{x^2} - 5x - 6$
As we see that the highest power of x is 2 so it is a quadratic equation, if the highest power was 3 then it is a cubic equation.
Now factorize the given quadratic equation, so factorize the multiplication of constant term and coefficient of ${x^2}$ such that the sum or difference gives us -5 and product gives us $\left( {6 \times - 6} \right) = - 36$
As the coefficient of ${x^2}$ is 6, coefficient of x is negative 5 and constant term is -6, so factors of multiplication of constant term and coefficient of ${x^2}$ i.e. $\left( {6 \times - 6} \right) = - 36$, such that the sum of the factors will give us the coefficient of x i.e. -5 and the product of the factors will give us $\left( {6 \times - 6} \right) = - 36$
So -36 can be factorize as -9 and 4,
So the sum of -9 and 4 is -5.
And the product of -9 and 4 is -36.
So the given quadratic equation is written as
$ \Rightarrow 6{x^2} - 9x + 4x - 6$
Now take 3x common from first two terms and 2 from last two terms we have,
$ \Rightarrow 3x\left( {2x - 3} \right) + 2\left( {2x - 3} \right)$
Now take (2x – 3) common we have,
$ \Rightarrow \left( {2x - 3} \right)\left( {3x + 2} \right)$
So this is the required factorization of the given quadratic equation.
Note:
We can also solve this quadratic equation directly by using the quadratic formula which is given as, $\left( {x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right)$, where a, b and c are the coefficients of ${x^2}$, x and constant term respectively, so on comparing, a = 6, b = -5 and c= -6, so simply substitute these values in the given equation and simplify we will get the values of x let the values of x be p and q, so the factors are (x – p) and (x – q) respectively.
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