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Factorise f(x) = 6x^3 -7x^2 -x +2 = 0

Try to find first root: f(1) = 6 - 7 -1 + 2 = 0, therefore x-1 is a root. Find quadratic by inspection: (x-1)( )= 6x^3 -7x^2 -x +2 (x-1)(6x^2 - x - 2) Factorise quadratic: (x-1)(2x+1)(3x-2) = 0
TD
Answered by Tutor40745 D. • Maths tutor
10407 Views

Solve the differential equation dy/dx = y/x(x + 1) , given that when x = 1, y = 1. Your answer should express y explicitly in terms of x.

Rearrange differential equation to get 1/x(x+1) dx = 1/y dy. Separate x side into partial fractions where 1/x(x+1) = 1/x - 1/(x+1). Integrate each side. Resulting equation involves natural logs. Substitute i...
AT
Answered by Alexander T. • Maths tutor
17693 Views

The expansion of (1+x)^4 is 1 + 4x +nx^2 + 4x^3 + x^4. Find the value of n. Hence Find the integral of (1+√y)^4 between the values 1 and 0 (one top, zero bottom).

Using Binomial expansion or Pascal's triangle, expand (1+x)^4 to get 1+4x+6x^2+4x^3+x^4. Then, by substituting √y for x, get 1 + 4y^1/2 + 6y +4y^3/2 +y^2. Then, using the rules of integration, the expansion ...
TD
Answered by Tutor41123 D. • Maths tutor
7230 Views

Factorise y^2 + 7y + 6

So here we have a quadratic equation because it has the structure a^2 + bx + c. In order to factorise a quadratic we first need to look for two numbers that we can multiply together to get 6 and add together...
VH
Answered by Victoria H. • Maths tutor
14813 Views

Express 4 sin(x) – 8 cos(x) in the form R sin(x-a), where R and a are constants, R >0 and 0< a< π/2

4 sin(x) – 8 cos(x)= Rsin(x-a) here use double angle formula 4 sin(x) – 8 cos(x)= Rsin(x)cos(a)-Rcos(x)sin(a) Rearrange so in same format as LHS 4 sin(x) – 8 cos(x)= Rcos(a)sin(x)-Rsin(a)cos(x) Equate terms ...
SE
Answered by Simon E. • Maths tutor
25262 Views