
How do you factor completely $ 10a{{x}^{2}}-23ax-5a $ ?
Answer
560.4k+ views
Hint: We can take a common from the equation and then it will be a quadratic equation. We will factor the equation, in the same way, we factor any quadratic equation $ a{{x}^{2}}+bx+c $. We will write bx as a sum of 2 terms such that the product of their coefficients will be equal to the product of a and c.
Complete step by step answer:
The given equation is $ 10a{{x}^{2}}-23ax-5a $
We can see there is one a in every term of the equation so we can take that a common
$ \Rightarrow 10a{{x}^{2}}-23ax-5a=a\left( 10{{x}^{2}}-23x-5 \right) $
Now we can see $ \left( 10{{x}^{2}}-23x-5 \right) $ is a quadratic equation if compare this to $ a{{x}^{2}}+bx+c $ we get the value of a is equal to 10, the value of b is -23 and the value of c is -5
Now for factorization we have to spilt -23x into $ \alpha x+\beta x $ such that product of $ \alpha $ and $ \beta $ is equal to product of a and b
So $ \alpha \beta =ab=-50 $
We will pair all the integer whose product is -50 take the one out which has sum equal to -23
All the pairs are (1,-50) , (2,-25) , (5,-10) , (10,-5) , (25,-2) and (50,-1)
The pair (2,-25) has equal to -23
We write -23x as $ 2x-25x $
$ \left( 10{{x}^{2}}-23x-5 \right) $ = $ 10{{x}^{2}}+2x-25x-5 $
We can take 2x common in first half of the equation and -5 in the second half of the equation
$ \Rightarrow 10{{x}^{2}}+2x-25x-5=2x\left( 5x+1 \right)-5\left( 5x+1 \right) $
Now we can take $ 5x+1 $ common
$ \Rightarrow 2x\left( 5x+1 \right)-5\left( 5x+1 \right)=\left( 2x-5 \right)\left( 5x+1 \right) $
So we can write $ \left( 10{{x}^{2}}-23x-5 \right) $ as $ \left( 2x-5 \right)\left( 5x+1 \right) $
So factorization of $ a\left( 10{{x}^{2}}-23x-5 \right) $ is $ a\left( 2x-5 \right)\left( 5x+1 \right) $
Note:
We can check whether the factorization is correct or not by putting any value of the variable for example if we put x as 1, 2, 3 in the equation $ 10a{{x}^{2}}-23ax-5a $ we will get -18a, -11a, and 16a. If we put x as 1, 2, 3 in the equation $ a\left( 2x-5 \right)\left( 5x+1 \right) $ we will get -18a, -11a, 16a
We are getting the same answer so we have done factorization correctly.
Complete step by step answer:
The given equation is $ 10a{{x}^{2}}-23ax-5a $
We can see there is one a in every term of the equation so we can take that a common
$ \Rightarrow 10a{{x}^{2}}-23ax-5a=a\left( 10{{x}^{2}}-23x-5 \right) $
Now we can see $ \left( 10{{x}^{2}}-23x-5 \right) $ is a quadratic equation if compare this to $ a{{x}^{2}}+bx+c $ we get the value of a is equal to 10, the value of b is -23 and the value of c is -5
Now for factorization we have to spilt -23x into $ \alpha x+\beta x $ such that product of $ \alpha $ and $ \beta $ is equal to product of a and b
So $ \alpha \beta =ab=-50 $
We will pair all the integer whose product is -50 take the one out which has sum equal to -23
All the pairs are (1,-50) , (2,-25) , (5,-10) , (10,-5) , (25,-2) and (50,-1)
The pair (2,-25) has equal to -23
We write -23x as $ 2x-25x $
$ \left( 10{{x}^{2}}-23x-5 \right) $ = $ 10{{x}^{2}}+2x-25x-5 $
We can take 2x common in first half of the equation and -5 in the second half of the equation
$ \Rightarrow 10{{x}^{2}}+2x-25x-5=2x\left( 5x+1 \right)-5\left( 5x+1 \right) $
Now we can take $ 5x+1 $ common
$ \Rightarrow 2x\left( 5x+1 \right)-5\left( 5x+1 \right)=\left( 2x-5 \right)\left( 5x+1 \right) $
So we can write $ \left( 10{{x}^{2}}-23x-5 \right) $ as $ \left( 2x-5 \right)\left( 5x+1 \right) $
So factorization of $ a\left( 10{{x}^{2}}-23x-5 \right) $ is $ a\left( 2x-5 \right)\left( 5x+1 \right) $
Note:
We can check whether the factorization is correct or not by putting any value of the variable for example if we put x as 1, 2, 3 in the equation $ 10a{{x}^{2}}-23ax-5a $ we will get -18a, -11a, and 16a. If we put x as 1, 2, 3 in the equation $ a\left( 2x-5 \right)\left( 5x+1 \right) $ we will get -18a, -11a, 16a
We are getting the same answer so we have done factorization correctly.
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