Question

How do you factor $64{x^3} + 1$?

Answer 1

Hint: If we have a quadratic equation then we have two methods for solving it one is the factoring method and another is the formula of Sreedhar Acharya. By using it we can solve our quadratic equation.
But for solving the third-degree equation or polynomial we have to solve it by putting value of the values one by one which value belongs to all real numbers like - $...... - 3, - 2, - 1,0,1,2,3,.....$.

Complete step by step Solution:
Given that –
Find the factor of $64{x^3} + 1$ polynomial
We know that the formula ${a^3} + {b^3} = (a + b)({a^2} + {b^2} - ab)$
Now according to the question, we need to solve the $64{x^3} + 1$
Now we can write given polynomial as ${(8x)^3} + {(1)^3}$
Now we will compare ${(8x)^3} + {(1)^3}$ with the ${a^3} + {b^3}$ then we will get the values of $a$ and $b$ which are $a = 8x$ and $b = 1$
Now we will put it in the formula of ${a^3} + {b^3} = (a + b)({a^2} + {b^2} - ab)$ then we will get
${(8x)^3} + {(1)^3} = (8x + 1)[{(8x)^2} + {(1)^2} - 8x \times 1]$
After multiplying all the values, we will get
${(8x)^3} + {(1)^3} = (8x + 1)[64{x^2} - 8x + 1]$
Now the one factor of $64{x^3} + 1$ is the $(8x + 1)$ and now we will find two more factor by solving the polynomial $64{x^2} - 8x + 1$
Now for solving the polynomial $64{x^2} - 8x + 1$ we will use the same method which we use for the solving the quadratic equation which is the Sreedhar Acharya formula then we will get 5 two more factor of polynomial $64{x^3} + 1$
Let - $64{x^2} - 8x + 1 = 0$ it will make easy to understand to finding the factor
Now we know that the formula of Sreedhar Acharya
If the general quadratic equation is the $a{x^2} + bx + c = 0$ then factor and solution of this equation is the
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now we will solve our given equation $64{x^2} - 8x + 1 = 0$
Now compare our given equation with the general form of equation $a{x^2} + bx + c = 0$ then
We will get the value of $a,b,c$
Now after comparison the value of $a = 64,b = - 8,c = 1$
Now put all value in the formula of Sreedhar Acharya $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ then
Now after putting all values we will get
$ \Rightarrow x = \dfrac{{ - ( - 8) \pm \sqrt {{{( - 8)}^2} - 4 \times 64 \times 1} }}{{2 \times 64}}$
After solving all the calculation, we will get our solution which is
$ \Rightarrow x = \dfrac{{ - ( - 8) \pm \sqrt {64 - 256} }}{{2 \times 64}}$
After solving the value of our above equation, we will get
$ \Rightarrow x = \dfrac{{ - ( - 8) \pm \sqrt {( - 192)} }}{{128}}$
Now we can write is as $x = \dfrac{{ - 8 \pm \sqrt { - (2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3)} }}{{128}}$
So, we will get three pair of $2$ so we will get $x = \dfrac{{ - 8 \pm 8\sqrt {( - 3)} }}{{128}}$
Now divide by $8$ we will get $x = \dfrac{{ - 1 \pm \sqrt {( - 3)} }}{{16}}$
So, we got two solution of $x$ one is in the positive form and another is in the form negative so our solution and factor of $x$ is the $x = \dfrac{{ - 1 + \sqrt {( - 3)} }}{{16}}$ and another is the $x = \dfrac{{ - 1 - \sqrt {( - 3)} }}{{16}}$

Therefore our cubic polynomial $64{x^3} + 1$ factor and solutions are $(8x + 1)$, $x = \dfrac{{ - 1 + \sqrt {( - 3)} }}{{16}}$ and $x = \dfrac{{ - 1 - \sqrt {( - 3)} }}{{16}}$which is our required answer of our question.

Note:
For solving any quadratic equation quickly we will use the factor method in which we have to find the factor of our given quadratic equation’s midterm (midterm is coefficient of variable $x$) such that their addition is equal to midterm and their multiply is equal to constant term, it is very easy and quick method for solving quadratic equation.