How do you factor #25x^2 - 10x + 4#?

1 Answer
Jun 24, 2016

Answer:

#25x^2-10x+4=(5x-1+isqrt3)(5x-1-isqrt3)#

Explanation:

As the polynomial #25x^2-10x+4# is quadratic, one can check discriminant to find if rational zeros and / or factors are possible or not. Discriminant of quadratic polynomial #(ax^2+bx+c)# is #(b^2-4ac)#. If it is negative, no real zeros / factors exist and if it is square of a number, rational zeros / factors are possible.

Here, #(b^2-4ac)=((-10)^2-4xx25xx4)=(100-400)=-300# and as such no further real zeros / roots exist, but we can find complex binomials as factors.

As if #alpha# and #beta# are zeros of #ax^2+bx+c#, then #ax^2+bx+c=a(x-alpha)(x-beta)# and hence let us find zeros of the trinomial using quadratic formula, which gives zeros as #(-b+-sqrt(b^2-4ac))/(2a)#. Hence, zeros of #25x^2-10x+4# are

#(-(-10)+-sqrt((-10)^2-4*25*4))/(2*25)=(10+-sqrt(-300))/50#

i.e. #(10+-10isqrt3)/50=(1+-isqrt3)/5#

Hence #25x^2-10x+4=25(x-(1-isqrt3)/5)(x-(1+isqrt3)/5)#

= #(5x-1+isqrt3)(5x-1-isqrt3)#