Question

What should be subtracted from the polynomial ${{x}^{2}}-16x+30$, so that 15 is the zero of the resulting polynomial?
A.30
B.14
C.15
D.16

Answer 1

Hint: For solving this problem, first we let the constant to be subtracted be c. Then, we assume another root to be k. Now, we obtain the sum and product of roots individually and then form two equations to solve for both c and k.

Complete step-by-step answer:
In algebra, a quadratic function is a polynomial function with one or more variables in which the highest-degree term is of the second degree. A single-variable quadratic function can be stated as:
$f(x)=a{{x}^{2}}+bx+c,\quad a\ne 0$
If we have two zeros of a quadratic equation then the polynomial could be formed by using the simplified result which could be stated as:
${{x}^{2}}-(a+b)x+ab$, where a and b are two zeroes of the equation.
According to our problem, the given equation is ${{x}^{2}}-16x+30$. Now, let c be added to the equation to obtain 3 as one of the zeroes. Therefore, the modified equation will be ${{x}^{2}}-16x+30-c=0$. Let the other zero be k.
Now, two equations can be formed by using the sum and product of zeroes as:
\[\begin{align}
  & a+b=16 \\
 & a=15,b=k \\
 & \therefore 15+k=16\ldots (1) \\
 & k=1 \\
 & ab=30-c \\
 & 15k=30-c\ldots (2) \\
\end{align}\]
From equation (1), we get k = 1. Putting k = 1 in equation (2), we get
$\begin{align}
  & 15\times 1=30-c \\
 & 30-c=15 \\
 & c=30-15 \\
 & c=15 \\
\end{align}$
Therefore, 15 must be subtracted from the polynomial.
Hence, option (c) is correct.
Note:This problem could be alternatively solved by using the concept of zeros of an equation. We are given that 15 is the zero of the final equation, so f (15) = 0. Satisfying 15 in the given polynomial ${{x}^{2}}-16x+30$, we get remainder as 15. Hence, 15 must be subtracted from the polynomial ${{x}^{2}}-16x+30$ to make 15 a factor of this equation.