Question

Factors of \[27 - 6x - {x^2}\] are:
A. $(3 - x)(9 + x)$
B. $(x - 3)(x + 9)$
C. $(x - 2)(x - 7)$
D. None of these

Answer 1

Hint: In order to do factorization of any quadratic expression first consider the constant term and factorize the constant term, and then consider the coefficient of the $x$ term, and then check if any of the factors of the constant term sums up to give the coefficient of the $x$ term and then simplify to factorize the expression.

Complete step-by-step solution:
The given quadratic expression is \[27 - 6x - {x^2}\], which can be re-written as \[ - {x^2} - 6x + 27\], now this is in the form of $a{x^2} + bx + c$.
Now consider \[ - {x^2} - 6x + 27\].
$ \Rightarrow - {x^2} - 6x + 27$
In factoring method, consider the constant term , here the constant is $27$, now factorize it.
Consider all the factors of $27$:
$ \Rightarrow 27 = 1 \times 27;$
$ \Rightarrow 27 = 3 \times 9;$
$ \Rightarrow 27 = 9 \times 3;$
$ \Rightarrow 27 = 27 \times 1;$
Now consider the term $ - 6x,$ as this term can be written as :
$ \Rightarrow - 6x = - 9x + 3x$;
Hence considering the factors of $27$which are $9$and $3$;
$\therefore $Re-writing the quadratic expression in factoring terms :
$ \Rightarrow - {x^2} - 6x + 27$
$ \Rightarrow - {x^2} - 9x + 3x + 27$
In this expression in first two terms taking $ - x$ term common and in the second two terms taking $3$term common:
$ \Rightarrow - x(x + 9) + 3(x + 9)$
$ \Rightarrow (3 - x)(x + 9)$
$\therefore - {x^2} - 6x + 27 = (3 - x)(x + 9)$
This can also be written as :
$\therefore - {x^2} - 6x + 27 = (3 - x)(9 + x)$

Option A is the correct answer.

Note: This is a polynomial of degree 2 as the highest power of the given polynomial is 2. As the degree of the polynomial is 2, it is a quadratic expression. The general form of a quadratic expression is given by $a{x^2} + bx + c$ , where here it is compared with \[27 - 6x - {x^2}\]. Here $a = - 1,b = - 6$ and $c = 27$.