How do you factor #6x^2 + 8x + 2#?

2 Answers
Don't Memorise
May 19, 2015

We can Split the Middle Term of this expression to factorise it
In this technique, if we have to factorise an expression like #ax^2 + bx + c#, we need to think of 2 numbers such that:
#N_1*N_2 = a*c = 6*2 = 12#
AND
#N_1 +N_2 = b = 8#
After trying out a few numbers we get #N_1 = 6# and #N_2 =2#
#6*2 = 12#, and #6+2= 8#

#6x^2 + 8x + 2 = 6x^2 + 6x + 2x + 2#

# = 6x(x+1) + 2(x+1)#

#(x+1)# is a common factor to each of the terms

# = (x+1)(6x+2)#

#=color(green)(2(x+1)(3x+1)#

Nghi N.
May 19, 2015

There is another shortcut that avoids the lengthy factoring by grouping.

#y = 2(3x^2 + 4x + 1).#

Since a + b + c = 0, the trinomial in parentheses has one factor (x + 1) and another (x + c/a) = (x + 1/3)

Factored form: #f(x) = 2(x + 1)(x + 1/3) = 2(x + 1)(3x + 1)#

Check by developing: #f(x) = 2(3x^2 + x + 3x + 1)# .OK

Reminder of the shortcut for f(x) = ax^2 + bx + c = 0.
1. When a + b + c = 0, one real roots is (1) and the other is (c/a)
2. When a - b + c = 0, one real root is (-1) and the other is (-c/a)

Remember this TIP. It will save you a lot of time and effort.

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