How do you factor completely 10ax^2 - 23ax - 5a10ax2−23ax−5a?
1 Answer
a(5x +1)(2x -5)
Explanation:
The first step is to remove the common factor a.
rArra(10x^2-23x-5)⇒a(10x2−23x−5) To factorise
10x^2-23x-510x2−23x−5 consider the following.The
color(blue)"standard form of the quadratic function"standard form of the quadratic function is
color(red)(|bar(ul(color(white)(a/a)color(black)(ax^2+bx+c)color(white)(a/a)|)))
and to factorise thisConsider the factors of the 'product' ac which also sum to give b, the coefficient of the x term.
here a = 10 , b = - 23 and c = -5
hence ac
=10xx-5=-50 The factors of -50 which also sum to give -23 are -25 and +2
Now express
10x^2-23x-5 as
10x^2+(color(red)"2x"-color(red)"25x")-5" [ -23x=2x-25x]" and factorising each 'pair' to obtain.
color(blue)"2x"(5x+1)-color(blue)"5"(5x+1) Taking out the common factor of (5x + 1)
(5x+1)(color(blue)"2x-5")
rArr10ax^2-23ax-5a=a(5x+1)(2x-5)
