Is it possible to factor #y=x^2+x-12 #? If so, what are the factors?

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Answer 1 · 2016-07-07T18:33:43

# y=(x-3)(x+4).#

Yes, #y=f(x), say, = x^2+x-12# can be factorised into two linear factors.

In fact, a quadratic polynomial like #ax^2+bx+c# can be factorised [ in RR], if and only if, #Delta=b^2-4ac>=0.#

In our case, we are having, #a=1, b= 1, c=-12,# so, #Delta=1-4*1*(-12)=1+48=49>=0#, hence, given #f# can be factorised.

Its factors are #(x-alpha), (x-beta)#, where, #alpha=( -b+sqrtDelta)/(2a)=(-1+sqrt49)/(2*1)=(-1+7)/2=3#

and, #beta=(-b-sqrtDelta0/(2a)=(-1-7)/2=-8/2=-4.

Hence the factors are #(x-3) and, (x+4).#

#:. y=(x-3)(x+4).#

Another Method, is, as follows :-

#y=f(x), say, = x^2+x-12#
#:. y=x^2+4x-3x-12.....................[4xx3=12, 4-3=1]#
#=x(x+4)-3(x+4)=(x+4)(x-3)#, as before!

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Answer 2 · 2016-07-07T19:06:20

#(x + 4)(x-3)#

Find two factors of 12 which differ (because of the MINUS 12) by 1, as indicated by the coefficient of the x term.

With a basic knowledge of the times tables we find:

# 4xx3 = 12 " and " +4 - 3 = +1#
These then are the factors we need. The MINUS 12 also indicates that the sign will be different, (neg x pos = neg)

#(x + 4)(x-3)#