Answer
385.8k+ views
Hint: This is a question of quadratic equations. The quadratic equation can be written as a multiplication of two linear equations. These linear equations represent the roots of the quadratic equation, one each. Factoring a quadratic equation means representing the equation in the multiplication of two linear functions. The general form of the quadratic equation as a product of two linear equation can be given as\[m\left( x-a \right)\left( x-b \right)\] where m is a coefficient of \[{{x}^{2}}\] and a, b are roots.
Complete step by step answer:
Here we can see the given equation is \[3{{x}^{2}}-8x=-4\] which can be rewritten as
\[3{{x}^{2}}-8x+4=0\]
We will solve the above equation using the Quadratic formula. The roots of a quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as
\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Comparing equation \[3{{x}^{2}}-8x=-4\]and \[a{{x}^{2}}+bx+c=0\]we get the parameters as
a= 3, b= -8 and c= 4
Substituting these values in the Quadratic formula we get
\[\Rightarrow \] \[\dfrac{-(-8)\pm \sqrt{{{8}^{2}}-4(3)(4)}}{2(3)}\]
Solving the expression inside square root we get
\[\begin{align}
& \\
& \Rightarrow \dfrac{8\pm \sqrt{64-48}}{6} \\
\end{align}\]
\[\Rightarrow \dfrac{8\pm \sqrt{16}}{6}\]
Now on solving further and putting \[\sqrt{16}=4\], we get
\[\Rightarrow \]\[\dfrac{8\pm 4}{6}\]
Now we know that the two roots are given as \[\dfrac{8+4}{6}\] and \[\dfrac{8-4}{6}\].
Solving further we get
\[\dfrac{12}{6},\dfrac{4}{6}\]
Hence the roots of the equation are \[2,\dfrac{2}{3}\] ( by reducing the fractions).
Now as we know the factors of the equation can be given as \[m\left( x-2 \right)\left( x-\dfrac{2}{3} \right)\], where m is the coefficient of \[{{x}^{2}}\].
From seeing this equation, \[3{{x}^{2}}-8x+4=0\], m can be given as 3.
Hence factor of \[3{{x}^{2}}-8x+4=0\] is \[3\left( x-2 \right)\left( x-\dfrac{2}{3} \right)\] or \[\left( x-2 \right)\left( 3x-2 \right)\].
Note:
Be attentive while calculating and substituting values of a, b, and c in the Quadratic formula.
The factors of the quadratic equation can also be found using the completing squares form.
Where the equation is converted into factors by forming it into the form \[{{a}^{2}}-2ab+{{b}^{2}}\]and then factors are given as (a+b)(a-b).
Complete step by step answer:
Here we can see the given equation is \[3{{x}^{2}}-8x=-4\] which can be rewritten as
\[3{{x}^{2}}-8x+4=0\]
We will solve the above equation using the Quadratic formula. The roots of a quadratic equation \[a{{x}^{2}}+bx+c=0\] is given as
\[\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Comparing equation \[3{{x}^{2}}-8x=-4\]and \[a{{x}^{2}}+bx+c=0\]we get the parameters as
a= 3, b= -8 and c= 4
Substituting these values in the Quadratic formula we get
\[\Rightarrow \] \[\dfrac{-(-8)\pm \sqrt{{{8}^{2}}-4(3)(4)}}{2(3)}\]
Solving the expression inside square root we get
\[\begin{align}
& \\
& \Rightarrow \dfrac{8\pm \sqrt{64-48}}{6} \\
\end{align}\]
\[\Rightarrow \dfrac{8\pm \sqrt{16}}{6}\]
Now on solving further and putting \[\sqrt{16}=4\], we get
\[\Rightarrow \]\[\dfrac{8\pm 4}{6}\]
Now we know that the two roots are given as \[\dfrac{8+4}{6}\] and \[\dfrac{8-4}{6}\].
Solving further we get
\[\dfrac{12}{6},\dfrac{4}{6}\]
Hence the roots of the equation are \[2,\dfrac{2}{3}\] ( by reducing the fractions).
Now as we know the factors of the equation can be given as \[m\left( x-2 \right)\left( x-\dfrac{2}{3} \right)\], where m is the coefficient of \[{{x}^{2}}\].
From seeing this equation, \[3{{x}^{2}}-8x+4=0\], m can be given as 3.
Hence factor of \[3{{x}^{2}}-8x+4=0\] is \[3\left( x-2 \right)\left( x-\dfrac{2}{3} \right)\] or \[\left( x-2 \right)\left( 3x-2 \right)\].
Note:
Be attentive while calculating and substituting values of a, b, and c in the Quadratic formula.
The factors of the quadratic equation can also be found using the completing squares form.
Where the equation is converted into factors by forming it into the form \[{{a}^{2}}-2ab+{{b}^{2}}\]and then factors are given as (a+b)(a-b).
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