Question

How do you factor \[{y^2} - 4y + 4\]?

Answer 1

Hint: A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. If factors are difficult to find then we use Sridhar’s formula to find the roots. That is \[y = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].

Complete step-by-step solution:
Given, \[{y^2} - 4y + 4\]
Since the degree of the equation is 2, we have 2 factors.
On comparing the given equation with the standard quadratic equation\[a{y^2} + by + c\], we have\[a = 1\], \[b = - 4\] and \[c = 4\].
The standard form of the factorization of quadratic equation is \[a{y^2} + {b_1}y + {b_2}y + c\], which satisfies the condition \[{b_1} \times {b_2} = a \times c\] and \[{b_1} + {b_2} = b\].
We can write the given equation as \[{y^2} - 2y - 2y + 4\], where \[{b_1} = - 2\] and \[{b_2} = - 2\]. Also \[{b_1} \times {b_2} = ( - 2) \times ( - 2) = 4(ac)\] and \[{b_1} + {b_2} = - 2 - 2 = - 4(b)\].
Thus we have,
\[ \Rightarrow {y^2} - 4y + 4 = {y^2} - 2y - 2y + 4\]
\[ = {y^2} - 2y - 2y + 4\]
Taking ‘y’ common in the first two terms and taking \[ - 2y\] common in the remaining two terms we have,
\[ \Rightarrow y(y - 2) - 2(y - 2)\]
Again taking \[(y - 2)\] common we have,
\[ = (y - 2)(y - 2)\]
Hence the factors of \[{y^2} - 4y + 4\] are \[(y - 2)\] and \[(y - 2)\]

Additional information:
If we know the algebraic identity \[{(a - b)^2} = {a^2} - 2ab + {b^2}\] we can solve this problem directly and easily.
Compared with the given problem that is \[{y^2} - 4y + 4\]. We have \[a = y\] and \[b = 2\].
\[ \Rightarrow {y^2} - 4y + 4 = {(y - 2)^2}\]
\[ \Rightarrow (y - 2)(y - 2)\]. In both cases we have the same answer.

Thus the factors of the given question are \[ = (y - 2)(y - 2)\]

Note: We can also find the roots of the given quadratic equation by equating the obtained factors to zero. That is
\[(y - 2)(y - 2) = 0\]
By zero product principle we have,
\[(y - 2) = 0\] or \[(y - 2) = 0\]
\[ \Rightarrow y = 2\] and \[y = 2\].

These are the roots of the given polynomial of degree 2. In above, if we are unable to expand the middle term of the given equation into a sum of two numbers then we use a quadratic formula to solve the given problem. Quadratic formula and Sridhar’s formula are both the same. Careful in the calculation part.