Question

Show that (x + 4), (x – 3) and (x – 7) are factors of ${{x}^{3}}-6{{x}^{2}}-19x+84$.

Answer 1

Hint: If $x+\alpha $ is factor of any polynomial f(x) $\to x=-\alpha $ will be a root of the same polynomial f(x) i.e. on putting $x=-\alpha \to $ f(x), f(x) will become 0. Use this property of roots of factors to solve the problem.

Complete step-by-step solution -
As we know, the factor of any polynomial will divide the polynomial completely, i.e. the remainder will be 0. And the factor of any polynomial gives the roots of the polynomial by equating the factor to 0. Let us suppose we have any polynomial f(x) which will give it’s roots by the equation
f(x) = 0 ………………… (i)
Now, let $\left( x+\alpha \right)$be the factor of f (x), then we can divide f(x)$\to \left( x+\alpha \right)$ and let the quotient be g(x). so, we can write f(x) as
$f\left( x \right)=\left( x+\alpha \right)\left( g\left( x \right) \right)$ ………….. (ii)
Now, from equation (i) and (ii), we get
$\begin{align}
  & \left( x+\alpha \right)\left( g\left( x \right) \right)=0 \\
 & x+\alpha =0,g\left( x \right)=0 \\
 & x=-\alpha ,g\left( x \right)=0 \\
\end{align}$
So, we can get roots of f(x) as $x=-\alpha $ and roots of equation g(x) = 0.
Hence, if $\left( x+\alpha \right)$ is a factor of f(X), then $x=-\alpha $ will be the root of f(x).
Now, coming to the question, we need to prove the factors of
${{x}^{3}}-6{{x}^{2}}-19x+84\to $ x + 4, x – 3, x – 7.
So, we have polynomial as
 $f\left( x \right)={{x}^{3}}-6{{x}^{2}}-19x+84$ ………………….(iii)
Now, let us verify whether the given factors are of given polynomial or not.
So, the first factor is given as x + 4. It means x = - 4 will be a root of the polynomial of equation (iii). So, it should be a solution of the given polynomial. So, let us put x = - 4 to equation (iii), we get
$f\left( -4 \right)={{\left( -4 \right)}^{3}}-6{{\left( -4 \right)}^{2}}-19\left( -4 \right)+84$
= - 64 – 96 + 76 + 84
= - 160 + 160
f(- 4) = 0
So, we get that x = - 4 is satisfying the given polynomial, it means x + 4 will be a factor of the polynomial.
Second factor is given as (x – 3), so, x = 3 should be a root of the given polynomial. So, we can pu x = 3 to the equation (iii) and hence, we get
$f\left( 3 \right)={{\left( 3 \right)}^{3}}-6{{\left( 3 \right)}^{2}}-19\left( 3 \right)+84$
=27 – 54 – 57 + 84
= - 27 + 27
f(x) = 0
Hence, (x – 3) will also be a factor of the given polynomial, as x = 3 is satisfying it.
Last factor of the polynomial is given as (x -7), so, x = 7 should be a root of the given polynomial. Hence, put x = 7 to the equation (iii), we get
 $f\left( 7 \right)={{\left( 7 \right)}^{3}}-6{{\left( 7 \right)}^{2}}-19\left( 7 \right)+84$
= 343 – 294 -133 + 84
= 49 - 49
f(7) = 0
Hence, 7 is also a root of the given polynomial i.e. (x – 7) will be a factor of it.
Hence, it is shown that factors (x + 4), (x – 3), (x – 7) are factors pf ${{x}^{3}}-6{{x}^{2}}-19x+84$.

Note: Another approach for the given problem would be that we can factorize the given polynomial and hence compare with the given polynomial. As x = 3 is a root of the polynomial (by hit and trial), so, (x – 3) will be a factor and hence find other factors by dividing the whole expression by (x – 3).
Roots of the polynomial will be – 4, 3, 7 as per the given factors of the polynomial. One may go wrong if he/she takes the roots of the polynomial as 4, -3, -7 by getting confused with the factors,
x + 4, x – 3, x – 7.
So, don’t be confused with this step.