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f(x) = x^3+2x^2-x-2 . Solve for f(x) = 0

f(x) = x³ + 2x² - x - 2 Use the factor theorem to test for f(1) = 0 f(1) = 1³ + 2*1² - 1 - 2 = 0 Therefore x = 1 is a solution and (x-1) is a factor of f(x) Now...

Answered by Elena S. • Maths tutor

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If y=2x+4x^3+3x^4 and z=(1+x)^2, find dy/dx and dz/dx.

Differentiation dy/dx = 2+12x^2+12x^3
Chain ruleu = 1+xz=u^2dz/dx = dz/du * du/dx = 2u * 1 = 2(1+x)

Answered by • Maths tutor

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solve this simulatneous equations (with clear algebraic working) : 5x-2y = 33 , 5x + 8y = 18

5x -2y = 33 (substract) 5x + 8y = 18
-10y = 15
y = -1.5
5x + 8(-1.5) = 18
5x = 30
x= 6

Answered by Lara E. • Maths tutor

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What is the coefficient of x^4 in the expansion of (x+3)^7

Start by expanding (a+b)⁷, using Pascal's Triangle or the binomial coefficient function to work out the coefficients:a⁷ + 7a⁶b + 21a⁵b² + 35a⁴

Answered by Josef W. • Maths tutor

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Integrate 1/(1 - 3*x) with respect to x

First substitute: u = 1 - 3xNext calculate: du/dx = -3 .... therefore dx = (-1/3) * duNow re-arrange the expression: Integrate 1/u * ( -1/3)*duNext recall the integral of 1/x is the natural logarithm, and...

Answered by Neil M. • Maths tutor

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