Question

Is ${{x}^{2}}-10x+25$ a perfect square trinomial and how do you factor it?

Answer 1

Consider the coefficient of ${{x}^{2}}$, coefficient of x and the constant terms as a, b and c respectively. Now, calculate the discriminant (D) given as: - $D={{b}^{2}}-4ac$, of the given equation. If the discriminant value is 0 then we can say that the given quadratic polynomial is a perfect square trinomial. To factor the equation write it in the form ${{a}^{2}}-2ab+{{b}^{2}}$ and use the conversion ${{\left( a-b \right)}^{2}}$ to get the answer.

Complete answer:
Here, we have been provided with the quadratic polynomial: ${{x}^{2}}-10x+25$ and we are asked to check if it is a perfect square trinomial or not. In the next step we have to factorize it.
Now, the given quadratic polynomial will only be a perfect square trinomial if it can be written in the form ${{\left( x-m \right)}^{2}}$. As we can see that both the roots of the equation will be the same if such a condition arises, so the discriminant value must be zero. Considering the coefficient of ${{x}^{2}}$, coefficient of x and the constant term as a, b and c respectively, we have,
For ${{x}^{2}}-10x+25$,
$\Rightarrow$ a = 1
$\Rightarrow$ b = -10
$\Rightarrow$ c = 25
Applying the formula of discriminant (D), we have,

& \Rightarrow D={{b}^{2}}-4ac \\\ & \Rightarrow D={{\left( -10 \right)}^{2}}-4\times 1\times 25 \\\ & \Rightarrow D=100-100 \\\ & \Rightarrow D=0 \\\ \end{aligned}$$ Therefore, the value of discriminant turns out to be 0, so we can say that $${{x}^{2}}-10x+25$$ is a perfect square trinomial. Now, let us factor the given quadratic polynomial. We can write the given expression as: - $$\because {{x}^{2}}-10x+25={{x}^{2}}-2\times x\times 5+{{5}^{2}}$$ Clearly, the R.H.S. of the above expression is of the form $${{a}^{2}}-2ab+{{b}^{2}}$$ whose factored form is $${{\left( a-b \right)}^{2}}$$, so we can write, $$\Rightarrow {{x}^{2}}-10x+25={{\left( x-5 \right)}^{2}}$$ Hence, $${{\left( x-5 \right)}^{2}}$$ is the factored form of the given quadratic expression. **Note:** One may note that if the value of D would have been greater than 0 then we would have used the middle term split method or completing the square method to factorize the quadratic polynomial. If the value of D would have been negative, i.e., less than 0, then we cannot factorize the quadratic polynomial. If the value of D would have been negative, i.e., less than 0, then we cannot factorize the polynomial because in that case the roots would not have been real. Remember the formulas: - $${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right),{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}},{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$$, because they are used everywhere in the topic ‘algebra’.