
How do you write \[12{{x}^{3}}-4{{x}^{2}}\] in factored form?
Answer
557.7k+ views
Hint: If \[x=a\] is a root of a polynomial function, then \[\left( x-a \right)\] is a factor of the polynomial. To write a polynomial function in its factored form, we need to multiply all its factors. For a general cubic expression of the form \[a{{x}^{3}}+b{{x}^{2}}+cx+d\], its factored form will be of the form \[a\left( x-\alpha \right)\left( x-\beta \right)\left( x-\delta \right)\], here \[\alpha ,\beta \And \delta \] are the roots of the function. We can find one root of the expression by hit and trial method, after this we can express the cubic as the product of the factor of the root we found, and a quadratic expression. We can easily factorize the quadratic expression to find other factors.
Complete step by step answer:
The given expression is \[12{{x}^{3}}-4{{x}^{2}}\], comparing with the general form of the cubic expression \[a{{x}^{3}}+b{{x}^{2}}+cx+d\], here we have \[a=12\]. First, we have to find one root of the equation by hit and trial method.
Substitute \[x=0\] in the cubic expression, we get
\[\Rightarrow 12{{(0)}^{3}}-4{{(0)}^{2}}=0\]
Hence one root of the expression is \[x=0\]. So, we can express the given cubic expression as, \[12{{x}^{3}}-4{{x}^{2}}=(x-0)(12{{x}^{2}}-4x)\]
Now we need to factorize the quadratic \[12{{x}^{2}}-4x\], as we can see that this expression has no constant terms, so one of its root must be equals to zero,
Hence the factored form of the quadratic is \[12{{x}^{2}}-4x=(x-0)(12x-4)\]
We can use this factored form of the quadratic to express given cubic as factored form as follows,
\[\Rightarrow 12{{x}^{3}}-4{{x}^{2}}=(x-0)(12{{x}^{2}}-4x)\]
\[\Rightarrow 12{{x}^{3}}-4{{x}^{2}}=(x-0)(x-0)(12x-4)\]
Dividing and multiplying RHS of the above expression by 12, we get
\[\Rightarrow 12{{x}^{3}}-4{{x}^{2}}=(x-0)(x-0)(12x-4)\times \dfrac{12}{12}\]
\[\begin{align}
& \Rightarrow 12{{x}^{3}}-4{{x}^{2}}=12(x-0)(x-0)\dfrac{(12x-4)}{12} \\
& \Rightarrow 12{{x}^{3}}-4{{x}^{2}}=12(x-0)(x-0)\left( x-\dfrac{1}{3} \right) \\
\end{align}\]
The factored form of the given cubic expression is as above.
Note:
It should be noted that we can express a polynomial function as factored form only if it has real roots. For example, the quadratic expression \[{{x}^{2}}+1\] can not be expressed as a factored form, as it has no real roots.
Complete step by step answer:
The given expression is \[12{{x}^{3}}-4{{x}^{2}}\], comparing with the general form of the cubic expression \[a{{x}^{3}}+b{{x}^{2}}+cx+d\], here we have \[a=12\]. First, we have to find one root of the equation by hit and trial method.
Substitute \[x=0\] in the cubic expression, we get
\[\Rightarrow 12{{(0)}^{3}}-4{{(0)}^{2}}=0\]
Hence one root of the expression is \[x=0\]. So, we can express the given cubic expression as, \[12{{x}^{3}}-4{{x}^{2}}=(x-0)(12{{x}^{2}}-4x)\]
Now we need to factorize the quadratic \[12{{x}^{2}}-4x\], as we can see that this expression has no constant terms, so one of its root must be equals to zero,
Hence the factored form of the quadratic is \[12{{x}^{2}}-4x=(x-0)(12x-4)\]
We can use this factored form of the quadratic to express given cubic as factored form as follows,
\[\Rightarrow 12{{x}^{3}}-4{{x}^{2}}=(x-0)(12{{x}^{2}}-4x)\]
\[\Rightarrow 12{{x}^{3}}-4{{x}^{2}}=(x-0)(x-0)(12x-4)\]
Dividing and multiplying RHS of the above expression by 12, we get
\[\Rightarrow 12{{x}^{3}}-4{{x}^{2}}=(x-0)(x-0)(12x-4)\times \dfrac{12}{12}\]
\[\begin{align}
& \Rightarrow 12{{x}^{3}}-4{{x}^{2}}=12(x-0)(x-0)\dfrac{(12x-4)}{12} \\
& \Rightarrow 12{{x}^{3}}-4{{x}^{2}}=12(x-0)(x-0)\left( x-\dfrac{1}{3} \right) \\
\end{align}\]
The factored form of the given cubic expression is as above.
Note:
It should be noted that we can express a polynomial function as factored form only if it has real roots. For example, the quadratic expression \[{{x}^{2}}+1\] can not be expressed as a factored form, as it has no real roots.
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