Question

How do you factor by grouping \[{x^3} + 4{x^2} + 8x + 32\]?

Answer 1

Hint: Here we use hit and trial method to find one of the zeroes of the given polynomial. Use the concept of zeroes of a polynomial that they give polynomial value equal to zero when substituted in place of variables. Use the obtained value of zero to make one of the factors of the polynomial. Divide the given polynomial by the factor and calculate remaining factors.
* A polynomial of degree \[n\] is a polynomial where the variable has highest power \[n\]. It can be written as \[a{x^n} + b{x^{n - 1}} + ..... + cx + d = 0\]
* Any polynomial can have repetitive zeroes, i.e. the zeroes of a polynomial need not be different from each other.
* If \[x = a\] is one of the zeroes of a polynomial \[a{x^n} + b{x^{n - 1}} + ..... + cx + d = 0\] then we divide the polynomial by \[x - a\] to find other zeros.

Complete step-by-step solution:
We are given the polynomial \[{x^3} + 4{x^2} + 8x + 32\]
Let us write \[f(x) = {x^3} + 4{x^2} + 8x + 32\].............… (1)
Hit and trial method states that we put in that value in place of x which makes the polynomial as 0 by guessing.
Put \[x = - 1\]
\[ \Rightarrow f( - 1) = {( - 1)^3} + 4{( - 1)^2} + 8( - 1) + 32\]
\[ \Rightarrow f( - 1) = - 1 + 4 - 8 + 32\]
\[ \Rightarrow f( - 1) = 27\]
Put \[x = - 2\]
\[ \Rightarrow f( - 2) = {( - 2)^3} + 4{( - 2)^2} + 8( - 2) + 32\]
\[ \Rightarrow f( - 2) = - 8 + 16 - 16 + 32\]
\[ \Rightarrow f( - 2) = 24\]
Put \[x = - 3\]
\[ \Rightarrow f( - 3) = {( - 3)^3} + 4{( - 3)^2} + 8( - 3) + 32\]
\[ \Rightarrow f( - 3) = - 27 + 36 - 24 + 32\]
\[ \Rightarrow f( - 3) = 17\]
Put \[x = - 4\]
\[ \Rightarrow f( - 4) = {( - 4)^3} + 4{( - 4)^2} + 8( - 4) + 32\]
\[ \Rightarrow f( - 4) = - 64 + 64 - 32 + 32\]
\[ \Rightarrow f( - 4) = 0\]
We can see that if we substitute the value of x as -4 we get the polynomial as 0.
So we can write \[x = - 4\] when substituted in the polynomial gives the value zero.
\[ \Rightarrow \left( {x + 4} \right)\] is a factor of the polynomial \[{x^3} + 4{x^2} + 8x + 32\]
So, to find other zeros of the polynomial we divide the polynomial by \[\left( {x + 4} \right)\] .
Using long division method
\[x + 4)\overline {{x^3} + 4{x^2} + 8x + 32} ({x^2} + 8\]
          \[\underline {{x^3} + 4{x^2}} \]
          \[8x + 32\]
           \[\underline {8x + 32} \]
                     \[0\]
Therefore, from long division method we write
\[(x + 4).({x^2} + 8) = {x^3} + 4{x^2} + 8x + 32\]
So, \[\left( {{x^2} + 8} \right)\] is another factor of the polynomial
Now we solve for zeros of a quadratic equation \[{x^2} + 8\]
Write \[{x^2} + 8 = 0\]
Shift coefficient to right hand side of the equation
\[ \Rightarrow {x^2} = - 8\]
Take square root on both sides of the equation
\[ \Rightarrow \sqrt {{x^2}} = \sqrt { - 8} \]
Write \[8 = {2^2}.2\]
\[ \Rightarrow \sqrt {{x^2}} = \sqrt { - {2^2}.2} \]
Cancel square root by square power on both sides of the equation
\[ \Rightarrow x = 2\sqrt { - 2} \]
Now we can write \[\sqrt { - 2} = i\sqrt 2 \] as we know \[\sqrt { - 1} = i\]
As we get the solution as complex or imaginary values, we will not expand the factor in equation (2) further.

Factors of the polynomial \[{x^3} + 4{x^2} + 8x + 32\] are \[(x + 4)\] and \[\left( {{x^2} + 8} \right)\]

Note: Students are likely to make mistakes in the process of long division as they forget to change the sign before the next step. Also, note that a cubic polynomial can have at most three zeroes. Keep in mind, here we don’t write the value of factors using complex numbers as we have to write the polynomial using factor form i.e. multiplication of two or more factors, so we can stop after we get two factors.