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Hint: The values of length, breadth and height are the zeroes of the given cubic polynomial. So we have to find out the roots of the polynomial to get the required values.

Complete step-by-step answer:

Here it is given that the values of length, breadth and height are the zeroes of the polynomial: ${{x}^{3}}-6{{x}^{2}}+11x-6$

Zeros of a polynomial can be defined as the points where the polynomial becomes zero on the whole. They are also known as the roots of the polynomial.

Here the given polynomial is of degree three. Polynomial of degree three is known as a cubic polynomial. Cubic polynomials have three roots. We have to find out those roots.

There we have to solve the following equation:

${{x}^{3}}-6{{x}^{2}}+11x-6=0$

If we put $x=1$ on the left hand side of the equation we will get zero.

${{x}^{3}}-6{{x}^{2}}+11x-6={{\left( 1 \right)}^{3}}-6{{\left( 1 \right)}^{2}}+11-6=12-12=0$

That means 1 is a root of the given polynomial. If 1 is root that means $\left( x-1 \right)$ will be a factor of the polynomial. We can rearrange the polynomial as:

$\begin{align}

& {{x}^{3}}-6{{x}^{2}}+11x-6 \\

& ={{x}^{2}}\left( x-1 \right)-5x\left( x-1 \right)+6\left( x-1 \right) \\

\end{align}$

We have $\left( x-1 \right)$ with each of the terms. Therefore we can take it out as a common factor.

$=\left( x-1 \right)\left( {{x}^{2}}-5x+6 \right)$

$\begin{align}

& =\left( x-1 \right)\left( {{x}^{2}}-2x-3x+6 \right) \\

& =\left( x-1 \right)\left( x\left( x-2 \right)-3\left( x-2 \right) \right) \\

& =\left( x-1 \right)\left( x-2 \right)\left( x-3 \right) \\

\end{align}$

Therefore,

${{x}^{3}}-6{{x}^{2}}+11x-6=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$

Now we have,

$\begin{align}

& {{x}^{3}}-6{{x}^{2}}+11x-6=0 \\

& \Rightarrow \left( x-1 \right)\left( x-2 \right)\left( x-3 \right)=0 \\

\end{align}$

Hence the roots of the equation are,

$\begin{align}

& x-1=0\Rightarrow x=1 \\

& x-2=0\Rightarrow x=2 \\

& x-3=0\Rightarrow x=3 \\

\end{align}$

The three roots are 1, 2 and 3.

Therefore, the length, breadth and height are 1, 2 and 3 respectively.

Note: Alternatively, we can solve the equation by using the relation between roots and coefficients of the polynomial equation. We know that one root is 1.

Sum of the three roots = 6

Therefore sum of the other two roots $=6-1=5$.

And the product of the three roots is also 6. One of them is 1. Therefore the product of two roots is 6.

Hence the other two roots are 2 and 3.

Complete step-by-step answer:

Here it is given that the values of length, breadth and height are the zeroes of the polynomial: ${{x}^{3}}-6{{x}^{2}}+11x-6$

Zeros of a polynomial can be defined as the points where the polynomial becomes zero on the whole. They are also known as the roots of the polynomial.

Here the given polynomial is of degree three. Polynomial of degree three is known as a cubic polynomial. Cubic polynomials have three roots. We have to find out those roots.

There we have to solve the following equation:

${{x}^{3}}-6{{x}^{2}}+11x-6=0$

If we put $x=1$ on the left hand side of the equation we will get zero.

${{x}^{3}}-6{{x}^{2}}+11x-6={{\left( 1 \right)}^{3}}-6{{\left( 1 \right)}^{2}}+11-6=12-12=0$

That means 1 is a root of the given polynomial. If 1 is root that means $\left( x-1 \right)$ will be a factor of the polynomial. We can rearrange the polynomial as:

$\begin{align}

& {{x}^{3}}-6{{x}^{2}}+11x-6 \\

& ={{x}^{2}}\left( x-1 \right)-5x\left( x-1 \right)+6\left( x-1 \right) \\

\end{align}$

We have $\left( x-1 \right)$ with each of the terms. Therefore we can take it out as a common factor.

$=\left( x-1 \right)\left( {{x}^{2}}-5x+6 \right)$

$\begin{align}

& =\left( x-1 \right)\left( {{x}^{2}}-2x-3x+6 \right) \\

& =\left( x-1 \right)\left( x\left( x-2 \right)-3\left( x-2 \right) \right) \\

& =\left( x-1 \right)\left( x-2 \right)\left( x-3 \right) \\

\end{align}$

Therefore,

${{x}^{3}}-6{{x}^{2}}+11x-6=\left( x-1 \right)\left( x-2 \right)\left( x-3 \right)$

Now we have,

$\begin{align}

& {{x}^{3}}-6{{x}^{2}}+11x-6=0 \\

& \Rightarrow \left( x-1 \right)\left( x-2 \right)\left( x-3 \right)=0 \\

\end{align}$

Hence the roots of the equation are,

$\begin{align}

& x-1=0\Rightarrow x=1 \\

& x-2=0\Rightarrow x=2 \\

& x-3=0\Rightarrow x=3 \\

\end{align}$

The three roots are 1, 2 and 3.

Therefore, the length, breadth and height are 1, 2 and 3 respectively.

Note: Alternatively, we can solve the equation by using the relation between roots and coefficients of the polynomial equation. We know that one root is 1.

Sum of the three roots = 6

Therefore sum of the other two roots $=6-1=5$.

And the product of the three roots is also 6. One of them is 1. Therefore the product of two roots is 6.

Hence the other two roots are 2 and 3.

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