Question

How do you determine if x - 5 is a factor of $2{{x}^{3}}-4{{x}^{2}}-7x-10$ ?

Answer 1

Hint: In this problem, we have to divide the two polynomials by long division method. If the factor is satisfying the expression that means it gives the remainder zero when divided. In the long division method, we have to move step by step focussing on the leading term of both the polynomials. Here x – 5 is binomial and the other one is polynomial.

Complete step-by-step answer:
Now, let’s solve the question.
Polynomial is an algebraic expression having many terms. The highest power in the expression is called the degree of the polynomial. There are 3 different types of polynomials: monomial, binomial and trinomial. Monomial consists of only a single term. Whereas binomial consists of two terms. And the trinomial consists of 3 three terms. When it comes to division, we prefer a long division method for dividing any monomial with polynomial or binomial with a polynomial. We are given a case of dividing a binomial with a polynomial.
The expression x – 5 is a divisor whereas $2{{x}^{3}}-4{{x}^{2}}-7x-10$ is a dividend. After this, the leading term of the dividend is divided by the leading term of the divisor so that we can form the expression such that it gets easily divisible.
$\Rightarrow \dfrac{2{{x}^{3}}}{x}=2{{x}^{2}}$
Now, multiply $2{{x}^{2}}$ with the divisor.
 $\Rightarrow 2{{x}^{2}}(x-5)=2{{x}^{3}}-10{{x}^{2}}$
Now this term gets subtracted from the dividend. After subtraction, we will get a new dividend. Then again the same procedure will be followed. Let’s go step by step by performing it:
$x-5\overset{2{{x}^{2}}+6x+23}{\overline{\left){\begin{align}
  & 2{{x}^{3}}-4{{x}^{2}}-7x-10 \\
 & 2{{x}^{3}}-10{{x}^{2}} \\
 & \overline{\begin{align}
  & 0{{x}^{3}}+6{{x}^{2}}-7x-10 \\
 & \text{ }6{{x}^{2}}-30x \\
 & \overline{\text{ }0{{x}^{2}}+23x-10} \\
\end{align}} \\
 & \text{ }23x-115 \\
 & \overline{\text{ }0x+95} \\
\end{align}}\right.}}$
We can easily see from the above division that x – 5 is not the factor because it is not giving us the remainder as zero.

Note: Please note that the terms of the dividend and the divisor are arranged in decreasing order of their degrees before solving by long division method. There is a shortcut method of checking that the factor is completely divisible by the polynomial or not.
From this it is said that when a divisor is an exact factor of f(x), then the remainder after division will be 0.
If f(a) = 0 $\Leftrightarrow $ x – a is a factor of f(x). Now let f(x) = $2{{x}^{3}}-4{{x}^{2}}-7x-10$. Then equate:
$\Rightarrow x-5=0\Leftrightarrow x=5$
Put x = 5 in the equation:
$\Rightarrow 2{{\left( 5 \right)}^{3}}-4{{\left( 5 \right)}^{2}}-7\left( 5 \right)-10$
On solving further:
f(x) = 105
So, f(x) $\ne $ 0 which states that x – 5 is not a factor of $2{{x}^{3}}-4{{x}^{2}}-7x-10$.