
How do you factor the expression ${{x}^{2}}-4x+4$?
Answer
492k+ views
Hint: In this question we have a trinomial given which we will solve by splitting the middle term similar to how we solve a quadratic equation and then taking out the common terms and grouping them together and write them together such that no term exists which as a power in it.
Complete step by step answer:
We have the given equation as:
$\Rightarrow {{x}^{2}}-4x+4$
Now since the above equation is in the quadratic format, we will find the value of $x$ as by splitting the equation, the above equation can be split up as:
$\Rightarrow {{x}^{2}}-2x-2x+4$
Now the above equation can be grouped as:
$\Rightarrow x(x-2)-2(x-2)$
Since the term $(x-2)$ is same in both the terms, we can take it out as common and write it as:
$\Rightarrow (x-2)(x-2)$
Now since both the terms are the same, we can group them as:
$\Rightarrow {{(x-2)}^{2}}$, which is the factored format of the given equation, and is the required solution.
Note: It is to be noted that in the above question we don’t have a quadratic equation, but we are factoring the trinomial by considering it as a quadratic equation.
A trinomial is an algebraic equation which has three terms in it.
A quadratic equation is a polynomial equation with a degree $2$, quadratic equations are used mostly in statistics when there is a power.
It is to be remembered that to split the middle term two terms should be such that the product is ${{x}^{2}}$ coefficient times the constant coefficient, and the sum is equal to the $x$ coefficient.
The roots of a quadratic equation can be found using the formula $({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2ac}$
Where $({{x}_{1}},{{x}_{2}})$ are the roots of the equation and $a,b,c$ are the coefficients of the terms in the quadratic equation.
It is not necessary that both the roots of the equation will be the same and some quadratic equation might not have proper roots.
Complete step by step answer:
We have the given equation as:
$\Rightarrow {{x}^{2}}-4x+4$
Now since the above equation is in the quadratic format, we will find the value of $x$ as by splitting the equation, the above equation can be split up as:
$\Rightarrow {{x}^{2}}-2x-2x+4$
Now the above equation can be grouped as:
$\Rightarrow x(x-2)-2(x-2)$
Since the term $(x-2)$ is same in both the terms, we can take it out as common and write it as:
$\Rightarrow (x-2)(x-2)$
Now since both the terms are the same, we can group them as:
$\Rightarrow {{(x-2)}^{2}}$, which is the factored format of the given equation, and is the required solution.
Note: It is to be noted that in the above question we don’t have a quadratic equation, but we are factoring the trinomial by considering it as a quadratic equation.
A trinomial is an algebraic equation which has three terms in it.
A quadratic equation is a polynomial equation with a degree $2$, quadratic equations are used mostly in statistics when there is a power.
It is to be remembered that to split the middle term two terms should be such that the product is ${{x}^{2}}$ coefficient times the constant coefficient, and the sum is equal to the $x$ coefficient.
The roots of a quadratic equation can be found using the formula $({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2ac}$
Where $({{x}_{1}},{{x}_{2}})$ are the roots of the equation and $a,b,c$ are the coefficients of the terms in the quadratic equation.
It is not necessary that both the roots of the equation will be the same and some quadratic equation might not have proper roots.
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