Factorize: \[{x^3} - 3{x^2} - 9x - 5\]
VerifiedHint: Here, we will factorize the given equation using factor theorem. Factorization or factoring is defined as the breaking or decomposition of an entity into a product of another entity, or factors, which when multiplied together give the original number or a matrix, etc.
Complete step-by-step answer:
We are given a quadratic equation \[{x^3} - 3{x^2} - 9x - 5\].
Let \[f\left( x \right) = {x^3} - 3{x^2} - 9x - 5\]
According to the Factor Theorem, \[k\] is a zero of \[f\left( x \right)\] if and only if \[\left( {x - k} \right)\] is a factor of \[f\left( x \right)\].
Substituting \[x = 1\] in \[f\left( x \right)\], we have
\[ \Rightarrow f\left( 1 \right) = {1^3} - 3{\left( 1 \right)^2} - 9\left( 1 \right) - 5 = - 16 \ne 0\]
So, \[x - 1\] is not a factor of \[f\left( x \right)\].
Substituting \[x = - 1\] in \[f\left( x \right)\], we have
\[ \Rightarrow f\left( { - 1} \right) = {( - 1)^3} - 3{( - 1)^2} - 9( - 1) - 5\]
Applying the exponent, we get
\[ \Rightarrow f\left( { - 1} \right) = - 1 - 3 + 9 - 5\]
Adding and subtracting the terms, we get
\[ \Rightarrow f\left( { - 1} \right) = 9 - 9 = 0\]
So, is a factor of \[f\left( x \right)\].
Now, by using synthetic division, we have
So, Divisor\[ = {x^2} - 4x - 5\] and Remainder \[ = 0\].
Now by using the formula Dividend \[ = \]Quotient \[ \times \] Divisor \[ + \] Remainder, we can write\[{x^3} - 3{x^2} - 9x - 5 = (x + 1) \cdot ({x^2} - 4x - 5) + 0\]
\[ \Rightarrow {x^3} - 3{x^2} - 9x - 5 = (x + 1) \cdot ({x^2} - 4x - 5)\]
By factoring, we get
\[ \Rightarrow {x^3} - 3{x^2} - 9x - 5 = (x + 1) \cdot ({x^2} + x - 5x - 5)\]
Now taking common factors, we get
\[ \Rightarrow {x^3} - 3{x^2} - 9x - 5 = (x + 1) \cdot (x(x + 1) - 5(x + 1))\]
Factoring out common terms, we get
\[ \Rightarrow {x^3} - 3{x^2} - 9x - 5 = (x + 1)(x + 1)(x - 5)\]
\[ \Rightarrow {x^3} - 3{x^2} - 9x - 5 = {(x + 1)^2}(x - 5)\]
Thus the factors are \[{(x + 1)^2},\left( {x - 5} \right)\]
\[{x^3} - 3{x^2} - 9x - 5 = {(x + 1)^2}(x - 5)\]
Note: The above problem has been done by using the factor theorem. The factor theorem states that If \[f(x)\] is a polynomial of degree \[n \ge 1\] and \['a'\] is any real number then
(i)\[f(a) = 0\] implies \[(x - a)\] is a factor of \[f(x)\].
(ii)\[(x - a)\] is a factor of \[f(x)\] implies \[f(a) = 0\].
We can also find factors using remainder theorems. Remainder theorem states that If a polynomial \[f(x)\]of degree greater than or equal to one is divided by a linear polynomial \[(x - a)\] then the remainder is \[f(a)\], where \['a'\] is any real number.
Synthetic division is also a way to find the zeros of a polynomial. Synthetic division can be defined as a simplified way of dividing a polynomial with another polynomial equation of degree \[1\] and is generally used to find the zeros of polynomials
