How to factorise 4x^2-4x+1?

Don't see how I can factorise to ${\left(2 x - 1\right)}^{2}$? Thank you so much!

Apr 14, 2018

${\left(2 x - 1\right)}^{2}$

Explanation:

$\text{for a quadratic in "color(blue)"standard form}$

•color(white)(x)ax^2+bx+c ;a!=0

$\text{consider the factors of the product ac which sum to b}$

$4 {x}^{2} - 4 x + 1 \text{ is in standard form}$

$\text{with "a=4,b=-4" and } c = 1$

$\Rightarrow a c = 4 \times 1 = 4 \text{ and - 2 , - 2 sum to - 4}$

$\text{split the middle term using these factors}$

$4 {x}^{2} - 2 x - 2 x + 1 \leftarrow \textcolor{b l u e}{\text{factorise in groups}}$

$= \textcolor{red}{2 x} \left(2 x - 1\right) \textcolor{red}{- 1} \left(2 x - 1\right)$

$\text{take out the common factor } \left(2 x - 1\right)$

$= \left(2 x - 1\right) \left(\textcolor{red}{2 x - 1}\right)$

$\Rightarrow 4 {x}^{2} - 4 x + 1 = {\left(2 x - 1\right)}^{2}$

Apr 14, 2018

$4 {x}^{2} - 4 x + 1 = {\left(2 x - 1\right)}^{2}$ is a perfect square trinomial...

Explanation:

Given:

$4 {x}^{2} - 4 x + 1$

This is an example of a perfect square trinomial.

Let's take a look at what happens when you square a binomial using FOIL to help us:

${\left(A + B\right)}^{2} = \left(A + B\right) \left(A + B\right)$

$\textcolor{w h i t e}{{\left(A + B\right)}^{2}} = {\overbrace{A \cdot A}}^{\text{First"+overbrace(A * B)^"Outside"+overbrace(B * A)^"Inside"+overbrace(B * B)^"Last}}$

$\textcolor{w h i t e}{{\left(A + B\right)}^{2}} = {A}^{2} + A B + A B + {B}^{2}$

$\textcolor{w h i t e}{{\left(A + B\right)}^{2}} = {A}^{2} + 2 A B + {B}^{2}$

In our example, note that $4 {x}^{2} = {\left(2 x\right)}^{2}$ and $1 = {1}^{2}$ are both perfect squares. So we might think of putting $A = 2 x$ and $B = 1$. That would allow us to find:

${\left(2 x + 1\right)}^{2} = 4 {x}^{2} + 4 x + 1$

That gives us $+ 4 x$ instead of the $- 4 x$ that we want.

Note however that if we put $B = - 1$ instead of $B = 1$ then we still have ${B}^{2} = \left(- 1\right) \left(- 1\right) = 1$ as we need, but the middle term becomes $A B = \left(4 x\right) \left(- 1\right) = - 4 x$ as we also want.

So:

${\left(2 x - 1\right)}^{2} = {\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(- 1\right) + {\left(- 1\right)}^{2} = 4 {x}^{2} - 4 x + 1$

More generally, we can write:

${\left(A - B\right)}^{2} = {A}^{2} - 2 A B + {B}^{2}$

So given any quadratic in standard form, if the first and last terms are perfect squares and it matches the pattern ${A}^{2} + 2 A B + {B}^{2}$ or ${A}^{2} - 2 A B + {B}^{2}$, then you can recognise it as ${\left(A + B\right)}^{2}$ or ${\left(A - B\right)}^{2}$.