Factorise #76a^3+4b^3-45a^2b# ?
2 Answers
Irreducible factorization
Explanation:
Given:
Try to factor by finding the greatest common factor(GCF) of each term:
There is no GCF common to all 3 monomials.
Explanation:
Given:
#76a^3+4b^3-45a^2b#
Note that all of the terms of this polynomial in
To see why, see what happens when we divide through by
#(76a^3+4b^3-45a^2b)/a^3 = (76a^3)/a^3+(4b^3)/a^3-(45a^2b)/a^3#
#color(white)((76a^3+4b^3-45a^2b)/a^3) = 76+4(b/a)^3-45(b/a)#
#color(white)((76a^3+4b^3-45a^2b)/a^3) = 4(b/a)^3-45(b/a)+76#
#color(white)((76a^3+4b^3-45a^2b)/a^3) = 4x^3-45x+76" "# where#x = b/a#
So if we can factor
#76a^3+4b^3-45a^2b = 4b^3-45a^2b+76a^3#
Linear factors correspond to zeros, so let's see if we can find zeros:
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/4# ,#+-1/2# ,#+-1# ,#+-2# ,#+-4# ,#+-19/4# ,#+-19/2# ,#+-19# ,#+-38# ,#+-76#
Trying each in turn, we eventually find:
#f(-4) = 4(color(blue)(-4))^3-45(color(blue)(-4))+76#
#color(white)(f(-4)) = -256+180+76#
#color(white)(f(-4)) = 0#
So
#4x^3-45x+76 = (x+4)(4x^2-16x+19)#
with the remaining quadratic, note that it takes the form:
#ax^2+bx+c#
with
This has discriminant
#Delta = b^2-4ac = (color(blue)(-16))^2-4(color(blue)(4))(color(blue)(19)) = 256-304 = -48#
Since
So let's stop with:
#4x^3-45x+76 = (x+4)(4x^2-16x+19)#
and hence (putting
#4(b/a)^3-45(b/a)+76 = ((b/a)+4)(4(b/a)^2-16(b/a)+19)#
and hence (multiplying through by
#4b^3-45a^2b+76a^3 = (b+4a)(4b^2-16ab+19a^2)#