Question

How do you graph \[{x^2} + {y^2} = 25\] ?

Answer 1

Hint: To solve this we need to give the values of ‘x’ and we can find the values of ‘y’. Otherwise we can find the coordinate of the given equation lying on the line of x- axis, we can find this by substituting the value of ‘y’ is equal to zero (x-intercept). Similarly we can find the coordinate of the equation lying on the line of y- axis, we can find this by substituting the value of ‘x’ equal to zero (y-intercept).

Formula used: The formula used in here is that for the first factor, we simply put the value of a root and solve the equation to get a zero. If we get a zero then, \[\left( x-a \right)\]is a factor for the equation. Now after that, we divide the equation with the factor \[\left( x-a \right)\] to get a quotient that gives the second factor, that quotient is a quadratic equation that is further split into factors.

Complete step-by-step solution:
 Let us start solving the question by taking the equation,
\[{{x}^{3}}-4{{x}^{2}}-2x+8=0\]
Now, let us think of a root that completely satisfies this equation,
If
 \[\begin{align}
  & \Rightarrow x=4 \\
 & \Rightarrow {{4}^{3}}-4\centerdot {{4}^{2}}-2\left( 4 \right)+8=64-64-8+8=0 \\
\end{align}\]
So, we can say that the factor \[\left( x-4 \right)\]satisfies the equation, now dividing the equation\[{{x}^{3}}-4{{x}^{2}}-2x+8\]with\[\left( x-4 \right)\], we get
\[\left( x-4 \right){{\left| \!{\overline {\,
 \begin{align}
  & {{x}^{3}}-4{{x}^{2}}-2x+8 \\
 & _{-}{{x}^{3}}{{-}_{+}}4{{x}^{2}}\downarrow +\downarrow \\
 & \underline{0+0} \\
 & -2x+8 \\
 & \underline{_{+}-2x{{+}{+}_{-}}8} \\
 & =0 \\
\end{align} \,}} \right. }^{{{x}^{2}}-2}}\]
Therefore, we can say that
\[\left( x-4 \right)\left( {{x}^{2}}-2 \right)={{x}^{3}}-4{{x}^{2}}-2x+8\]
So, now, we just need to make the further factors for the factor \[\left( {{x}^{2}}-2 \right)\]
\[\begin{align}
  & \left( {{x}^{2}}-2 \right) \\
 & \Rightarrow \left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right) \\
\end{align}\]
Thus, we can write that the factors for the equation\[{{x}^{3}}-4{{x}^{2}}-2x+8\] are
\[\left( x-4 \right)\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)\]
Hence the equation has been factored.

Hence the factors of the given equation\[{{x}^{3}}-4{{x}^{2}}-2x+8\] are
\[\left( x-4 \right)\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)\]

Note: The given equation is a cubic equation, that is why we get three factors. However, as the factor \[\left( x-4 \right)\]is repeated twice, its square has been taken but the degree of the equation still remains the same i.e. \[3\]. The factors can be verified by further multiplying them to get the same equation as before.