Question

If two zeroes of the polynomial ${x^4} - 6{x^3} - 26{x^2} + 138x - 35$ are $2 \pm \sqrt 3$ find the other zeroes.

Answer 1

According to the question, firstly calculate a single factor using the given zeros of the polynomial. Then, divide the polynomial with the calculated factor. Thus, the quotient came and made the factors using splitting the middle term method.

Formula used:
Here, we use an algebraic identity to solve the factor that are ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ and ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.

Complete step by step answer:
Let us assume the polynomial be $f(x)$. So, $f(x) = {x^4} - 6{x^3} - 26{x^2} + 138x - 35$.
As, the given roots of the polynomial are: $2 \pm \sqrt 3$ that is $2 + \sqrt 3$ and $2 - \sqrt 3$.
As, from the given roots we can say that $x = 2 + \sqrt 3$ and $x = 2 - \sqrt 3$ are zeroes.
So, on simplifying both the zeroes we get $x - 2 - \sqrt 3$ and $x - 2 + \sqrt 3$ are factors of the given polynomial.
Hence, we will multiply both the factors to get a single factor that is: $\left( {x - 2 - \sqrt 3 } \right) \times \left( {x - 2 + \sqrt 3 } \right)$
Hence, it becomes an identity that is ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$. Here, a is $x - 2$ and b is $\sqrt 3$ .
By substituting the values of a and b in the identity we get, ${\left( {x - 2} \right)^2} - {\left( {\sqrt 3 } \right)^2}$
On solving the squares that is by using the identity ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ we get,
${x^2} + 4 - 4x - 3$
On simplifying the above equation we get ${x^2} - 4x + 1$ is a factor.
Now we will divide $f(x) = {x^4} - 6{x^3} - 26{x^2} + 138x - 35$ by ${x^2} - 4x + 1$ .
So that we can also find the other factors.
By using division algorithm we get,

\\!\\!\\!\\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^4} - 6{x^3} - 26{x^2} + 138x - 35\\\\{x^4} - 4{x^3} + {x^2}\\\\\dfrac{{( - ){\rm{ }}( + ){\rm{ }}( - ){\rm{ }}}}{{{\rm{ }}\dfrac{\begin{array}{l} - 2{x^3} - 27{x^2} + 138x - 35\\\ - 2{x^3} + 8{x^2}{\rm{ }} - 2x\\\\{\rm{ }}( + ){\rm{ }}( - ){\rm{ }}( + )\end{array}}{{{\rm{ }}\dfrac{\begin{array}{l} - 35{x^2}{\rm{ }} + 140x{\rm{ }} - 35\\\ - 35{x^2}{\rm{ }} + 140x{\rm{ }} - 35\end{array}}{0}}}}}{\rm{ }}\end{array}}}} \limits^{\displaystyle\,\,\, {{x^2} - 2x - 35}}$$ Now, we will find out the zeroes of the $${x^2} - 2x - 35$$ Therefore, $${x^2} - 2x - 35 = 0$$ Here, by using splitting the middle term method we calculate its zeroes. In this method we will find out two numbers whose sum is $$ - 2$$ and product is $$ - 35$$. So, we are getting the two numbers which are $$ - 7$$ and 5 . $${x^2} - 2x - 35 = 0$$ $${x^2} - 7x + 5x - 5 = 0$$ Taking out common in the pairs of 2 we get, $$x\left( {x - 7} \right) + 5\left( {x - 7} \right) = 0$$ Taking 2 same factors one time we get, $$\left( {x - 7} \right)\left( {x + 5} \right) = 0$$ By separating out 2 factors, $$x - 7 = 0$$ Here, on simplifying the above equation we get, So, $$x = 7$$ Similarly, we can calculate the second value of x that is $$x + 5 = 0$$ . So, $$x = - 5$$ Hence, we get $$x = 7, - 5$$ are the other zeros of the polynomial. **Therefore, the zeroes of the polynomial are $$2 + \sqrt 3 ,2 - \sqrt 3 , - 5$$and 7.** **Note:** To solve these types of questions, you must remember the division algorithm and calculate the zeros using the quotient only. You can solve the equation either by splitting the middle term method or by using the formula that is $$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$.