What should be added to the polynomial ${{x}^{2}}-5x+4$, so that 3 is the zero of the resulting polynomial?
A.1
B.2
C.4
D.5
Answer
642.6k+ views
Hint: For solving this problem, first we let the constant to be added as 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 ${{x}^{2}}-5x+4$. Now, let c be added to the equation to obtain 3 as one of the zeroes. Therefore, the modified equation will be ${{x}^{2}}-5x+4+c=0$. Let the other zero be k.
Now, two equations can be formed as:
$\begin{align}
& a+b=5 \\
& a=3,b=k \\
& 3+k=5\ldots (1) \\
& k=2 \\
& ab=4+c \\
& 3k=4+c\ldots (2) \\
\end{align}$
From equation (1), we get k = 2. Putting k = 2 in equation (2), we get
$\begin{align}
& 3\times 2=4+c \\
& 4+c=6 \\
& c=6-4 \\
& c=2 \\
\end{align}$
Therefore, 2 must be added to the polynomial.
Hence, option (b) is correct.
Note:This problem could be alternatively solved by using the concept of zeros of an equation. We are given that 3 is the zero of the final equation, so f (3) = 0. Satisfying 3 in the given polynomial ${{x}^{2}}-5x+4$, we get remainder as -2. Hence, 2 must be added to the polynomial ${{x}^{2}}-5x+4$ to make 3 a factor.
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 ${{x}^{2}}-5x+4$. Now, let c be added to the equation to obtain 3 as one of the zeroes. Therefore, the modified equation will be ${{x}^{2}}-5x+4+c=0$. Let the other zero be k.
Now, two equations can be formed as:
$\begin{align}
& a+b=5 \\
& a=3,b=k \\
& 3+k=5\ldots (1) \\
& k=2 \\
& ab=4+c \\
& 3k=4+c\ldots (2) \\
\end{align}$
From equation (1), we get k = 2. Putting k = 2 in equation (2), we get
$\begin{align}
& 3\times 2=4+c \\
& 4+c=6 \\
& c=6-4 \\
& c=2 \\
\end{align}$
Therefore, 2 must be added to the polynomial.
Hence, option (b) is correct.
Note:This problem could be alternatively solved by using the concept of zeros of an equation. We are given that 3 is the zero of the final equation, so f (3) = 0. Satisfying 3 in the given polynomial ${{x}^{2}}-5x+4$, we get remainder as -2. Hence, 2 must be added to the polynomial ${{x}^{2}}-5x+4$ to make 3 a factor.
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