How do you factor and solve \[6{{x}^{2}}-5x=6\]?
Answer
626.1k+ views
Hint: We solve this question using factorization method. We will split the middle term in a way that their sum is equal to the middle term and product is equal to product of first and last terms. Finally, we have to take terms as common and simplify them. we have to represent the equation as a product of \[\left( x-a \right)\left( x-b \right)\].
Complete step by step answer:
Here we have been asked to factorize the quadratic polynomial \[6{{x}^{2}}-5x=6\] by grouping.
First we have bring the whole equation into LHS side we will get
\[\Rightarrow 6{{x}^{2}}-5x-6\]
Now we have to split the middle term into two in such a way that the sum is equal to \[-5\] and the product is equal to the product of constant term \[6\]and \[6{{x}^{2}}\]that is \[36\].
To do this, first we need to prime factorize for the number \[36\]
So let us find prime factors for \[36\]
We can write \[36\] as
\[36=2\times 2\times 3\times 3\] as a product of primes.
Now we can take \[4\] and \[-9\] to satisfy the given condition.
\[\begin{align}
& -36=-9\times 4 \\
& -5=-9+4 \\
\end{align}\]
So these values satisfy the condition now we will split the middle term as \[-9x+4x\]
Then the polynomial will look like
\[\Rightarrow 6{{x}^{2}}-9x+4x-6\]
Now we can take common terms out to make them as factors.
Here we can see from the first two terms we can take \[3x\] as common and from the next two terms we can \[2\] as common . After taking the common terms out the equation will look like
\[\Rightarrow 3x\left( 2x-3 \right)+2\left( 2x-3 \right)\]
Now in the above polynomial we can see \[2x-3\] as common and we can take it out then the polynomial will look like
\[\Rightarrow \left( 3x+2 \right)\left( 2x-3 \right)\]
Now the polynomial is represented as a product of factors.
So the factors of the polynomial \[6{{x}^{2}}-5x-6\] are \[\left( 3x+2 \right)\left( 2x-3 \right)\].
Now we can solve these factors for getting \[x\] values also.
We will get
\[\Rightarrow 3x=-2\]
\[\Rightarrow 2x=3\]
By simplifying these we will get
\[\Rightarrow x=\dfrac{-2}{3}\]
\[\Rightarrow x=\dfrac{3}{2}\]
So the values of \[x\] are \[\dfrac{-2}{3}\] and \[\dfrac{3}{2}\].
Note: We can also solve the above question using discriminant method also. In which we can solve the \[x\] for values a and b after that making them factors as \[\left( x-a \right)\left( x-b \right)\]. Here we will find factors. First there we will find values of \[x\].
Complete step by step answer:
Here we have been asked to factorize the quadratic polynomial \[6{{x}^{2}}-5x=6\] by grouping.
First we have bring the whole equation into LHS side we will get
\[\Rightarrow 6{{x}^{2}}-5x-6\]
Now we have to split the middle term into two in such a way that the sum is equal to \[-5\] and the product is equal to the product of constant term \[6\]and \[6{{x}^{2}}\]that is \[36\].
To do this, first we need to prime factorize for the number \[36\]
So let us find prime factors for \[36\]
We can write \[36\] as
\[36=2\times 2\times 3\times 3\] as a product of primes.
Now we can take \[4\] and \[-9\] to satisfy the given condition.
\[\begin{align}
& -36=-9\times 4 \\
& -5=-9+4 \\
\end{align}\]
So these values satisfy the condition now we will split the middle term as \[-9x+4x\]
Then the polynomial will look like
\[\Rightarrow 6{{x}^{2}}-9x+4x-6\]
Now we can take common terms out to make them as factors.
Here we can see from the first two terms we can take \[3x\] as common and from the next two terms we can \[2\] as common . After taking the common terms out the equation will look like
\[\Rightarrow 3x\left( 2x-3 \right)+2\left( 2x-3 \right)\]
Now in the above polynomial we can see \[2x-3\] as common and we can take it out then the polynomial will look like
\[\Rightarrow \left( 3x+2 \right)\left( 2x-3 \right)\]
Now the polynomial is represented as a product of factors.
So the factors of the polynomial \[6{{x}^{2}}-5x-6\] are \[\left( 3x+2 \right)\left( 2x-3 \right)\].
Now we can solve these factors for getting \[x\] values also.
We will get
\[\Rightarrow 3x=-2\]
\[\Rightarrow 2x=3\]
By simplifying these we will get
\[\Rightarrow x=\dfrac{-2}{3}\]
\[\Rightarrow x=\dfrac{3}{2}\]
So the values of \[x\] are \[\dfrac{-2}{3}\] and \[\dfrac{3}{2}\].
Note: We can also solve the above question using discriminant method also. In which we can solve the \[x\] for values a and b after that making them factors as \[\left( x-a \right)\left( x-b \right)\]. Here we will find factors. First there we will find values of \[x\].
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