Question

How do you factor and solve \[{y^2} + 8y = 0?\]

Answer 1

Hint: In this question we have to find the roots from the above quadratic equation by difference of cubes identity. Use the addition axiom to write the equation in standard form. Next, factor out all common factors and set each factor containing a variable equal to zero using the zero - product property. Solve for the variable in each equation formed from the above step. Finally, check the solution in the original equation.

Complete step by step answer:
Given:
\[ \Rightarrow {y^2} + 8y = 0\]
Next, take y in the above equation and we get
\[ \Rightarrow y(y + 8) = 0\]
Now, separate the above equation and we get
$ \Rightarrow y = 0$ and $y + 8 = 0$
$ \Rightarrow y = 0$ and $y = - 8$

This is the required factor of the given equation. y = 0 and y = - 8

Note: We have to remember that, for factoring polynomials, "factoring" (or” factoring completely”) is always done using some set of numbers as possible coefficient.
Here, \[{x^2} - 5\] can’t be factored using integer coefficients. (It is irreducible over the integers.)
In fact, in general we find:
\[ \Rightarrow {a^n} - {b^n} = (a - b)({a^n}^{ - 1} + {a^n}^{ - 2}b + {a^n}^{ - 3}{b^2} + ... + {b^n}^{ - 1})\]
Next we multiply the out in the above term. Most of the terms cancel. When n is even, the second factor can be factor further:
\[ \Rightarrow (a - b)({a^n}^{ - 1} + {a^n}^{ - 2}b + {a^n}^{ - 3}{b^2} + ... + {b^n}^{ - 1}) = \left( {a + b} \right)\left( {{a^n}^{ - 2} + {a^n}^{ - 4}{b^2} + {b^n}^{ - 2}} \right)\]
There are no more linear factors with Real coefficients, but there are quadratic factors with Real coefficients.
There are so far 8 common methods to solve quadratic equations, they are graphing, completing the squares, quadratic formula, factoring FOIL, the diagonal sum method, the popular factoring AC method, and the new transforming method.
When the quadratic equation can’t be factored, the quadratic formula is the obvious choice. However, solving by formula feels boring and repeating. In addition, the school math curriculum wants students to learn, beyond the formula, a few other solving methods. That is why many quadratic equations given in problems/tests/exams are internationally set up so that students have to solve them by other solving methods.