Question

Factorise `76a^3+4b^3-45a^2b` ?

Answer 1

Irreducible factorization

Explanation:

Given: `76a^3 + 4b^3 - 45a^2b`

Try to factor by finding the greatest common factor(GCF) of each term:

`76a^3 = 2 * 2 * 19 * a * a * a`

`4b^3 = 2 * 2 * b * b * b`

`- 45a^2b = -1 * 5 * 3 * 3 * a * a * b`

There is no GCF common to all 3 monomials.

Answer 2

`76a^3+4b^3-45a^2b = (b+4a)(4b^2-16ab+19a^2)`

Explanation:

Given:

`76a^3+4b^3-45a^2b`

Note that all of the terms of this polynomial in `a` and `b` are of degree `3`. So we can factor this as we might factor a cubic in a single variable.

To see why, see what happens when we divide through by `a^3`:

`(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 `f(x) = 4x^3-45x+76` then 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 `4x^3-45x+76` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `76` and `q` a divisor of the coefficient `4` of the leading term.

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 `x=-4` is a zero and `(x+4)` a factor:

`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 `a=4`, `b=-16` and `c=19`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (color(blue)(-16))^2-4(color(blue)(4))(color(blue)(19)) = 256-304 = -48`

Since `Delta < 0`, this quadratic has no real zeros and no linear factors with real coefficients.

So let's stop with:

`4x^3-45x+76 = (x+4)(4x^2-16x+19)`

and hence (putting `x=b/a`):

`4(b/a)^3-45(b/a)+76 = ((b/a)+4)(4(b/a)^2-16(b/a)+19)`

and hence (multiplying through by `a^3`):

`4b^3-45a^2b+76a^3 = (b+4a)(4b^2-16ab+19a^2)`