Answer:
Step-by-step explanation:
In the same way as you could factor trinomials on the form of
x2+bx+c
You can factor polynomials on the form of
ax2+bx+c
If a is positive then you just proceed in the same way as you did previously except now
ax2+bx+c=(x+m)(ax+n)
wherec=mn,ac=pqandb=p+q=am+n
Example
3x2−2x−8
We can see that c (-8) is negative which means that m and n does not have the same sign. We now want to find m and n and we know that the product of m and n is -8 and the sum of m and n multiplied by a (3) is b (-2) which means that we're looking for two factors of -24 whose sum is -2 and we also know that one of them is positive and of them is negative.
Factorsof−24−1,241,−24−2,122,−12−3,83,−8−4,64,−6Sumoffactors23−2310−105−52−2
This means that:
3x2−2x−8=
=3x2+(4−6)x−8=
=3x2+4x−6x−8
We can then group those terms that have a common monomial factor. The first two terms have x together and the second two -2 and then factor the two groups.
=(3x2+4x)+(−6x−8)=
=x(3x+4)−2(3x+4)
Notice that both remaining parenthesis are the same. This means that we can rewrite this using the distributive propertyit as:
=(x−2)(3x+4)=3x2−2x−8
This method is called factor by grouping.
A polynomial is said to be factored completely if the polynomial is written as a product of unfactorable polynomials with integer coefficients.
