Factorise the following: 5a^3b^5-4ab^2

Step 1:-Find any common factors. Looking at this example, it becomes clear that a HCF won't be a numerical value (e.g. 4 or 5) as there are no factors which are common for both 4 and 5. Looking further, can we use the a's and b's as factors? In this case, yes we can. The highest common factor in this case therefore is ab^2.
Step 2:-Using this found highest common factor, and removing (dividing) both equations by this, what are we left with?in this case, 5a^3b^5 will go to, ab^2(5a^2b^3), and the -4ab^2 will go to, ab^2(-4)
Step 3:-Putting this all together, we are left with the answer of:
ab^2(5a^2b^3-4)

Related Maths A Level answers

All answers ▸

A circle has the equation x^2 + y^2 - 4x + 10y - 115 = 0. Express the equation in the form (x - a)^2 + (y - b)^2 = k, and find the centre and radius of the circle.


Given that y = (1 + 3x^2)^(1/3) , use the chain rule to find dy/dx in terms of x.


integrate( x^3+4x^2+3)dx


The quadratic equation (k+1)x^2 + (5k - 3)x + 3k = 0 has equal roots. Find the possible values of k