How do you factor $2{{x}^{2}}-6x-8$ ?
Answer
616.2k+ views
Hint: On considering the given polynomial we can directly take $2$ common out and solve or else solve it without taking it out. Now since there are three terms, we must split the middle term which is $-6x$ into two terms so that we can factorize from the initial two terms and the final two terms. Now group them together to represent them as the factors for the polynomial.
Complete step by step solution:
The given polynomial which must be factorized is $2{{x}^{2}}-6x-8$
To factorize this polynomial, we must first split the middle term.
For this, we use the product sum form.
If the polynomial’s general form is $a{{x}^{2}}+bx+c=0$
Then to split the middle term, we use the product and sum formula.
Which is,
The product of the middle terms after splitting must be $a\times c$
And the sum of the middle terms must be $b$
Here $a=2;b=-6;c=-8$
The product of the terms is $2\times -8=-16$
The sum of the terms is $-6$
Therefore, the terms can be $2x,-8x$
On substituting back, we get,
$\Rightarrow 2{{x}^{2}}+2x-8x-8$
The polynomial is of degree $2$
Now, firstly let us take the common terms out of the first two terms.
$\Rightarrow 2x\left( x+1 \right)-8x-8$
Now secondly take the common terms out of the last two terms.
$\Rightarrow 2x\left( x+1 \right)-8\left( x+1 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow \left( 2x-8 \right)\left( x+1 \right)$
We can take $2$ common from the factor $2x-8$
Now writing all the factors together we get,
$\Rightarrow 2\left( x-4 \right)\left( x+1 \right)$
Hence the factors for the polynomial $2{{x}^{2}}-6x-8$ are $2\left( x-4 \right)\left( x+1 \right)$
Note: The process of factorization is the reverse multiplication. So, we have tried to get the question as a solution by multiplying all the factors together.
The factors are, $2\left( x-4 \right)\left( x+1 \right)$
On multiplying all of them we get,
$\Rightarrow \left( 2x-8 \right)\left( x+1 \right)$
$\Rightarrow \left( 2x\left( x+1 \right)-8\left( x+1 \right) \right)$
Now open the brackets to expand the terms,
$\Rightarrow \left( 2{{x}^{2}}+2x-8x-8 \right)$
Now evaluate,
$\Rightarrow \left( 2{{x}^{2}}-6x-8 \right)$
Since we got the question as the solution, our factors are correct.
Complete step by step solution:
The given polynomial which must be factorized is $2{{x}^{2}}-6x-8$
To factorize this polynomial, we must first split the middle term.
For this, we use the product sum form.
If the polynomial’s general form is $a{{x}^{2}}+bx+c=0$
Then to split the middle term, we use the product and sum formula.
Which is,
The product of the middle terms after splitting must be $a\times c$
And the sum of the middle terms must be $b$
Here $a=2;b=-6;c=-8$
The product of the terms is $2\times -8=-16$
The sum of the terms is $-6$
Therefore, the terms can be $2x,-8x$
On substituting back, we get,
$\Rightarrow 2{{x}^{2}}+2x-8x-8$
The polynomial is of degree $2$
Now, firstly let us take the common terms out of the first two terms.
$\Rightarrow 2x\left( x+1 \right)-8x-8$
Now secondly take the common terms out of the last two terms.
$\Rightarrow 2x\left( x+1 \right)-8\left( x+1 \right)$
Now on writing it in the form of the product of sums, also known as factoring,
$\Rightarrow \left( 2x-8 \right)\left( x+1 \right)$
We can take $2$ common from the factor $2x-8$
Now writing all the factors together we get,
$\Rightarrow 2\left( x-4 \right)\left( x+1 \right)$
Hence the factors for the polynomial $2{{x}^{2}}-6x-8$ are $2\left( x-4 \right)\left( x+1 \right)$
Note: The process of factorization is the reverse multiplication. So, we have tried to get the question as a solution by multiplying all the factors together.
The factors are, $2\left( x-4 \right)\left( x+1 \right)$
On multiplying all of them we get,
$\Rightarrow \left( 2x-8 \right)\left( x+1 \right)$
$\Rightarrow \left( 2x\left( x+1 \right)-8\left( x+1 \right) \right)$
Now open the brackets to expand the terms,
$\Rightarrow \left( 2{{x}^{2}}+2x-8x-8 \right)$
Now evaluate,
$\Rightarrow \left( 2{{x}^{2}}-6x-8 \right)$
Since we got the question as the solution, our factors are correct.
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