How to factorise 4x^2-4x+14x2−4x+1?
Don't see how I can factorise to (2x-1)^2(2x−1)2 ? Thank you so much!
Don't see how I can factorise to
2 Answers
Explanation:
"for a quadratic in "color(blue)"standard form"for a quadratic in standard form
•color(white)(x)ax^2+bx+c ;a!=0∙xax2+bx+c;a≠0
"consider the factors of the product ac which sum to b"consider the factors of the product ac which sum to b
4x^2-4x+1" is in standard form"4x2−4x+1 is in standard form
"with "a=4,b=-4" and "c=1with a=4,b=−4 and c=1
rArrac=4xx1=4" and - 2 , - 2 sum to - 4"⇒ac=4×1=4 and - 2 , - 2 sum to - 4
"split the middle term using these factors"split the middle term using these factors
4x^2-2x-2x+1larrcolor(blue)"factorise in groups"4x2−2x−2x+1←factorise in groups
=color(red)(2x)(2x-1)color(red)(-1)(2x-1)=2x(2x−1)−1(2x−1)
"take out the common factor "(2x-1)take out the common factor (2x−1)
=(2x-1)(color(red)(2x-1))=(2x−1)(2x−1)
rArr4x^2-4x+1=(2x-1)^2⇒4x2−4x+1=(2x−1)2
Explanation:
Given:
4x^2-4x+14x2−4x+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:
(A+B)^2 = (A+B)(A+B)(A+B)2=(A+B)(A+B)
color(white)((A+B)^2) = overbrace(A * A)^"First"+overbrace(A * B)^"Outside"+overbrace(B * A)^"Inside"+overbrace(B * B)^"Last"
color(white)((A+B)^2) = A^2+AB+AB+B^2
color(white)((A+B)^2) = A^2+2AB+B^2
In our example, note that
(2x+1)^2 = 4x^2+4x+1
That gives us
Note however that if we put
So:
(2x-1)^2 = (2x)^2+2(2x)(-1)+(-1)^2 = 4x^2-4x+1
More generally, we can write:
(A-B)^2 = A^2-2AB+B^2
So given any quadratic in standard form, if the first and last terms are perfect squares and it matches the pattern