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

Question

Eli Assouline · Accepted Answer

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

Avery Alexander · Answer

undefined To factor the expression 25x^2 - 10x + 4, you can use the quadratic formula:

$ ax^2 + bx + c = 0 $

where $ a = 25 $, $ b = -10 $, and $ c = 4 $.

The quadratic formula is:

$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $

Plugging in the values:

$ x = \frac{{-(-10) \pm \sqrt{{(-10)^2 - 4(25)(4)}}}}{{2(25)}} $

$ x = \frac{{10 \pm \sqrt{{100 - 400}}}}{{50}} $

$ x = \frac{{10 \pm \sqrt{{-300}}}}{{50}} $

Since the discriminant ($ b^2 - 4ac $) is negative, the roots will be complex. Therefore, the expression cannot be factored further using real numbers.