
Factorize the given quadratic polynomial: $ 2{{x}^{2}}-18 $ ?
Answer
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Hint:
We start solving the problem by adding and subtracting the term $ 6x $ to the given quadratic polynomial $ 2{{x}^{2}}-18 $ . We then make the necessary arrangements and make use of the distributive property $ a\left( b+c \right)=ab+ac $ to proceed through the problem. We then make use of the distributive property $ \left( a+b \right)c=ac+bc $ to get the factorization of the given quadratic polynomial which is the required answer.
Complete step by step answer:
According to the problem, we are asked to factorize the given quadratic polynomial: $ 2{{x}^{2}}-18 $.
Now, we have the polynomial $ 2{{x}^{2}}-18 $ .
Let us add and subtract the given polynomial with the term $ 6x $ .
$ \Rightarrow 2{{x}^{2}}-18=2{{x}^{2}}+6x-6x-18 $ .
$ \Rightarrow 2{{x}^{2}}-18=\left( 2x\times x \right)+\left( 2x\times 3 \right)+\left( -6\times x \right)+\left( -6\times 3 \right) $ ---(1).
From distributive property, we know that $ a\left( b+c \right)=ab+ac $ . Let us use this result in equation (1).
$ \Rightarrow 2{{x}^{2}}-18=2x\left( x+3 \right)-6\left( x+3 \right) $ ---(2).
From distributive property, we know that $ \left( a+b \right)c=ac+bc $ . Let us use this result in equation (2).
$ \Rightarrow 2{{x}^{2}}-18=\left( 2x-6 \right)\left( x+3 \right) $ .
So, we have found the factorization of the quadratic polynomial $ 2{{x}^{2}}-18 $ as $ \left( 2x-6 \right)\left( x+3 \right) $ .
$ \therefore $ The factors of the quadratic polynomial $ 2{{x}^{2}}-18 $ is $ \left( 2x-6 \right)\left( x+3 \right) $ .
Note:
We can also solve this problem by first finding the zeroes of the polynomial and then making use of the fact that the factors of the polynomial with zeroes a, b is $ \left( x-a \right) $ , $ \left( x-b \right) $ . We should keep in mind that if we multiply the obtained factors, we should get a similar polynomial as the result. We should not report the result of factorization as $ \left( x-3 \right)\left( x+3 \right) $ by dividing 2 from the factor which is the common mistake done by students. Similarly, we can expect problems to find the zeroes of the given quadratic polynomial.
We start solving the problem by adding and subtracting the term $ 6x $ to the given quadratic polynomial $ 2{{x}^{2}}-18 $ . We then make the necessary arrangements and make use of the distributive property $ a\left( b+c \right)=ab+ac $ to proceed through the problem. We then make use of the distributive property $ \left( a+b \right)c=ac+bc $ to get the factorization of the given quadratic polynomial which is the required answer.
Complete step by step answer:
According to the problem, we are asked to factorize the given quadratic polynomial: $ 2{{x}^{2}}-18 $.
Now, we have the polynomial $ 2{{x}^{2}}-18 $ .
Let us add and subtract the given polynomial with the term $ 6x $ .
$ \Rightarrow 2{{x}^{2}}-18=2{{x}^{2}}+6x-6x-18 $ .
$ \Rightarrow 2{{x}^{2}}-18=\left( 2x\times x \right)+\left( 2x\times 3 \right)+\left( -6\times x \right)+\left( -6\times 3 \right) $ ---(1).
From distributive property, we know that $ a\left( b+c \right)=ab+ac $ . Let us use this result in equation (1).
$ \Rightarrow 2{{x}^{2}}-18=2x\left( x+3 \right)-6\left( x+3 \right) $ ---(2).
From distributive property, we know that $ \left( a+b \right)c=ac+bc $ . Let us use this result in equation (2).
$ \Rightarrow 2{{x}^{2}}-18=\left( 2x-6 \right)\left( x+3 \right) $ .
So, we have found the factorization of the quadratic polynomial $ 2{{x}^{2}}-18 $ as $ \left( 2x-6 \right)\left( x+3 \right) $ .
$ \therefore $ The factors of the quadratic polynomial $ 2{{x}^{2}}-18 $ is $ \left( 2x-6 \right)\left( x+3 \right) $ .
Note:
We can also solve this problem by first finding the zeroes of the polynomial and then making use of the fact that the factors of the polynomial with zeroes a, b is $ \left( x-a \right) $ , $ \left( x-b \right) $ . We should keep in mind that if we multiply the obtained factors, we should get a similar polynomial as the result. We should not report the result of factorization as $ \left( x-3 \right)\left( x+3 \right) $ by dividing 2 from the factor which is the common mistake done by students. Similarly, we can expect problems to find the zeroes of the given quadratic polynomial.
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