Question

How do you factor ${(x + 2)^2} - 5(x + 2)$?

Answer 1

Hint: In the question, we have an equation that is quadratic, and we have to find its factors, which can be done with the help of rules. Factorisation is the reverse of multiplying out. One important factorisation process is the reverse of multiplications. The factors of any equation can be an integer, a variable or an algebraic expression itself.
There are many rules and formulas that help in factorising equations; we just have to observe which condition the equation satisfies.
Here, we have similarities with, ${(a + b)^2} = {a^2} + 2ab + {b^2}$ and we can find roots of the equation after that.

Complete step-by-step solution:
When we start solving this equation ${(x + 2)^2} - 5(x + 2)$,
First, we expand ${(x + 2)^2}$ which looks like \[(x + 2)(x + 2)\]
Next, we just multiply it out which will give us
$ \Rightarrow {x^2} + 4x + 4$
Now that we know,${(x + 2)^2}$ is equal to ${x^2} + 4x + 4$. We can do the other part which is \[5(x + 2)\]
$ \Rightarrow = 5x + 10$
Now we know \[5(x + 2) = (5x + 10)\], now we can write the problem out expanded,
$ \Rightarrow {x^2} + 4x + 4 - (5x + 10)$
Simplifying the equation, NOTICE THE NEGATIVE SIGN IN FRONT OF THE PARENTHESES
We have to distribute this negative to all terms in the parentheses, which gives us,
$ \Rightarrow {x^2} + 4x + 4 - 5x - 10$
Now we just combine like terms,
$ \Rightarrow {x^2} - x - 6$
Now we need 2 numbers that multiply to $ - 6$ and add up to $ - 1$
These numbers are $ - 3\,\,and\,\,2$
Notice $ - 3 \times 2 = - 6$
And $ - 3 + 2 = - 1$
So, we then get,
\[ \Rightarrow (x - 3)(x + 2)\]

Therefore, for an equation ${(x + 2)^2} - 5(x + 2)$ the factors will be \[(x - 3)(x + 2)\]

Note: In Mathematics, factorisation or factoring is defined as the breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity, or factors, which when multiplied together gives the original number or a matrix, etc. It is simply the resolution of an integer or polynomial into factors such that when multiplied together they will result in an original or initial integer or polynomial. In the factorisation method, we reduce any algebraic or quadratic equation into its simpler form, where the equations are represented as the product of factors instead of expanding the brackets.