Question

How do you factor \[y = {x^3} + {x^2} + 2x - 4\] ?

Answer 1

Hint: In this question, we have to solve the right-hand side of the given algebraic equations to get the factor.
First, we need to factor the constant term of the right-hand side. Then, putting these factors in the equation by trial-and-error method, we will get one of the roots as well as the factor. After that, put the value and apply the quadratic formula. We will get the other roots and the required solution.

Formula used:
The roots of the quadratic equation \[a{x^2} + bx + c = 0\] is
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step answer:
It is given that, \[y = {x^3} + {x^2} + 2x - 4\] .
We need to factor \[{x^3} + {x^2} + 2x - 4\] .
Since the coefficient of the constant term is \[4\] , then the possible factors are \[1,2,4\].
Now apply the factor theorem to check the possible values by trial-and-error method.
\[y = f(1) = {\left( 1 \right)^3} + {\left( 1 \right)^2} + 2 \times 1 - 4 = 1 + 1 + 2 - 4 = 0\]
\[y = f(2) = {\left( 2 \right)^3} + {\left( 2 \right)^2} + 2 \times 2 - 4 = 8 + 4 + 4 - 4 \ne 0\]
\[y = f(4) = {\left( 4 \right)^3} + {\left( 4 \right)^2} + 2 \times 4 - 4 = 64 + 16 + 8 - 4 \ne 0\]
Hence, \[x = 1\] is the first root.
Therefore, we can write the equation \[y = {x^3} + {x^2} + 2x - 4\] as,
\[y = {x^3} - {x^2} + 2{x^2} - 2x + 4x - 4\]
Or, \[y = {x^2}\left( {x - 1} \right) + 2x\left( {x - 1} \right) + 4\left( {x - 1} \right)\]
Or, \[y = \left( {x - 1} \right)\left( {{x^2} + 2x + 4} \right)\]
Now, we need to factor the term \[\left( {{x^2} + 2x + 4} \right)\] , using quadratic formula we can find out the factors of the quadratic equation \[{x^2} + 2x + 4 = 0\]
\[x = \dfrac{{ - 2 \pm \sqrt {{2^2} - 4 \times 1 \times 4} }}{2}\]
Or, \[x = \dfrac{{ - 2 \pm \sqrt {4 - 16} }}{2}\]
Or, \[x = \dfrac{{ - 2 \pm \sqrt { - 12} }}{2}\]
Simplifying we get,
\[x = \dfrac{{ - 2 \pm \sqrt {12} i}}{2}\] [since \[i = \sqrt { - 1} \] ]
Or, \[x = \dfrac{{ - 2 \pm 2\sqrt 3 i}}{2}\]
Or, \[x = - 1 \pm \sqrt 3 i\]
Then the two factors can be written as,
\[x - \left( { - 1 \pm \sqrt 3 i} \right)\]
i.e., \[\left( {x + 1 - \sqrt 3 i} \right),\left( {x + 1 + \sqrt 3 i} \right)\]
Then we can write, \[\left( {{x^2} + 2x + 4} \right) = \left( {x + 1 - \sqrt 3 i} \right)\left( {x + 1 + \sqrt 3 i} \right)\]
Therefore, we get,
\[y = \left( {x - 1} \right)\left( {x + 1 - \sqrt 3 i} \right)\left( {x + 1 + \sqrt 3 i} \right)\]

Hence, after factorization we can write,\[y = {x^3} + {x^2} + 2x - 4 = \left( {x - 1} \right)\left( {x + 1 - \sqrt 3 i} \right)\left( {x + 1 + \sqrt 3 i} \right)\].

Note:
> Algebraic expression:
In mathematics, an algebraic expression is an expression built up from integer constants, variables and the algebraic operations.
For example, \[{x^2} + 6xy + 7\] is an algebraic expression where 7 is the integer constants and x and y are the variables, + is the algebraic operations.
> Factor theorem:
In algebra, the factor theorem is a theorem linking factors and zeros of a polynomial.
The factor theorem states that, a polynomial \[f\left( x \right)\] has a factor \[\left( {x - a} \right)\] if and only if \[f\left( a \right) = 0\](i.e.\[a\] is a root).
i.e., let \[x = a\] is a solution of \[f\left( x \right)\], that could make \[\left( {x - a} \right)\] a factor.