Question

What should be added to the polynomial ${x^2} - 5x + 4$, so that $\left( {x - 3} \right)$ is a factor of the resulting polynomial?

Answer 1

Hint: In order to find the value needed to add in the polynomial to make it a factor of $\left( {x - 3} \right)$, we must know what factor and polynomial is. A Polynomial is an expression that contains more than one algebraic term, variables having different powers. A factor of a number is basically a number that divides the number completely without leaving any remainder.

Complete answer:
We are given with a polynomial ${x^2} - 5x + 4$, considering it to be $f\left( x \right)$, numerically written as:
$f\left( x \right) = {x^2} - 5x + 4$ ……. (1)
We are given with a value $\left( {x - 3} \right)$. Equating it with zero to get the zero of the polynomials and we get:
$\left( {x - 3} \right) = 0$
Adding both sides by 3:
$ \Rightarrow x - 3 + 3 = 0 + 3$
$ \Rightarrow x = 3$
Therefore, $f\left( x \right) = f\left( 3 \right)$ becomes the zero of the polynomials.
Since, we know that a factor is a number that completely divides a number and gets the remainder as zero.
And, we are given that if we add a number then $\left( {x - 3} \right)$ becomes the factor of the polynomial.
Let’s consider the number to be added as ‘p’:
So, the equation becomes:
\[f\left( x \right) = {x^2} - 5x + 4 + p = 0\]
Substitute $f\left( x \right) = f\left( 3 \right)$ in the above equation, and we get:
\[ \Rightarrow f\left( 3 \right) = {\left( 3 \right)^2} - 5\left( 3 \right) + 4 + p = 0\]
Solving it further:
\[ \Rightarrow f\left( 3 \right) = 9 - 15 + 4 + p = 0\]
\[ \Rightarrow f\left( 3 \right) = 13 - 15 + p = 0\]
\[ \Rightarrow f\left( 3 \right) = - 2 + p = 0\]
Adding both sides by \[2\], we get:
\[ \Rightarrow f\left( 3 \right) = - 2 + p + 2 = 0 + 2\]
\[ \Rightarrow p = 2\]
So, the number we add to the polynomial is \[2\], so that $\left( {x - 3} \right)$ is a factor of the resulting polynomial.
And, the resulting polynomial becomes:
$f\left( x \right) = {x^2} - 5x + 4 + 2$
$ \Rightarrow f\left( x \right) = {x^2} - 5x + 6$

Note:
1.We could have also solved this problem using the long division method by dividing the polynomial ${x^2} - 5x + 4$ by $\left( {x - 3} \right)$.
2.The zero of a polynomial is a value of x, that makes the value of $f\left( x \right)$ to be zero. For example, for $\left( {x - a} \right)$, “$a$” will be the value of x that makes $\left( {x - a} \right) = 0$ .
3.We can cross check the value $f\left( x \right) = {x^2} - 5x + 6$ by substituting $f\left( x \right) = f\left( 3 \right)$, to show $\left( {x - 3} \right)$ is a factor of the polynomial now.