Question

Solve the given polynomial: ${{x}^{2}}-5x+4$ .

Answer 1

Hint: We have to convert the given polynomial to the form $\left( x-a \right)\left( x-b \right)$ using completing the square method by representing it in terms of squares of numbers.

Complete step-by-step answer:
The quadratic polynomial given in the question is of the form $p\left( x \right)=a{{x}^{2}}+bx+c$. We have to convert this into the form $\left( x-\alpha \right)\left( x-\beta \right)$ where $\alpha $ and $\beta $ are the factors or zeroes of the given quadratic polynomial. We can convert polynomials to the above form by using the square method. In the square method, we convert the terms into squares of a particular term or number. Let us see how this can be achieved. The quadratic polynomial given in the question is: -
$p\left( x \right)={{x}^{2}}-5x+4$
Now, we are going to convert $p\left( x \right)$ in the form of ${{\left( x-a \right)}^{2}}-{{\left( b \right)}^{2}}$. Thus, after expanding, we will get,
${{\left( x-a \right)}^{2}}-{{\left( b \right)}^{2}}={{x}^{2}}-2ax+{{a}^{2}}-{{b}^{2}}$ ……………………………………(i)
Now, we will compare equation (i) with the quadratic polynomial given in question i.e. $p\left( x \right)$. After comparing we will get: -
$-2a=-5$
$\Rightarrow a=\dfrac{5}{2}$
$\Rightarrow {{a}^{2}}-{{b}^{2}}=4$
$\Rightarrow {{\left( \dfrac{5}{2} \right)}^{2}}-{{b}^{2}}=4$
$\Rightarrow {{b}^{2}}={{\left( \dfrac{5}{2} \right)}^{2}}-4$
$\Rightarrow {{b}^{2}}=\dfrac{9}{4}$
Now, using the values of a and ${{b}^{2}}$, we will convert $p\left( x \right)$ into desired form as shown below:-
$p\left( x \right)={{\left( x-\dfrac{5}{2} \right)}^{2}}-\dfrac{9}{4}$
Now, to obtain its zeroes i.e. $p\left( x \right)=0$ will give two values $\alpha $ and $\beta $ which are the zeroes of the given polynomial. Thus,
$p\left( x \right)=0$
$\Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}-\dfrac{9}{4}=0$
$\Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\dfrac{9}{4}={{\left( \dfrac{3}{2} \right)}^{2}}$
$\Rightarrow {{\left( x-\dfrac{5}{2} \right)}^{2}}=\pm {{\left( \dfrac{3}{2} \right)}^{2}}$
$\Rightarrow x=4$ and $x=1$
Thus, the zeroes of the above $p\left( x \right)$ are 4 and 1 respectively.
Now, we have to write $p\left( x \right)$ in terms of factors. Thus,
$p\left( x \right)=\left( x-4 \right)\left( x-1 \right)$
Thus, our required answer is $\left( x-4 \right)\left( x-1 \right)$.

Note: We could have also solved the above question with the help of quadratic formula. The quadratic formula is given by: -
$x=\dfrac{-b\pm D}{2a}$,
Where $D=\sqrt{{{b}^{2}}-4ac}$and x is the zeroes of the quadratic equation $a{{x}^{2}}+bx+c$.