By John Bird

A pragmatic creation to the center arithmetic rules required at: larger engineering point: John chicken s method of arithmetic, in line with various labored examples and: interactive difficulties, is perfect for vocational scholars that require an: complicated textbook: conception is stored to a minimal, with the emphasis firmly put on: problem-solving talents, making this a completely functional creation to: the complex arithmetic engineering that scholars have to grasp. The: vast and thorough subject assurance makes this a great textual content for top: point vocational classes: Now in its 7th variation, Engineering arithmetic has helped hundreds of thousands of: scholars to reach their assessments. the hot variation encompasses a part at: the beginning of every bankruptcy to give an explanation for why the content material is critical and the way: it pertains to actual existence. it's also supported by means of a completely up-to-date significant other

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E. the remainder is 3(1)3 + 2(1)2 + (−1)(1) + 4 = 3+2−1+4= 8 Similarly, when (x 3 − 7x − 6) is divided by (x − 3), the remainder is 1(3)3 + 0(3)2 − 7(3) − 6 = 0, which means that (x − 3) is a factor of (x 3 − 7x − 6) Here are some more examples on the remainder theorem. Problem 30. Without dividing out, find the remainder when 2x 2 − 3x + 4 is divided by (x − 2) 13 (i) Dividing (x 3 − 2x 2 − 5x + 6) by (x 2 + x − 2) gives: x −3 ————————– x 2 + x − 2 x 3 − 2x 2 − 5x + 6 x 3 + x 2 − 2x —————— −3x 2 − 3x + 6 −3x 2 − 3x + 6 ——————– · · · ——————– By the remainder theorem, the remainder is given by ap2 + bp + c, where a = 2, b = −3, c = 4 and p = 2.

Determine the remainder when x 3 − 6x 2 + x − 5 is divided by (a) (x + 2) (b) (x − 3) Section A Algebra Section A 14 Higher Engineering Mathematics 3. Use the remainder theorem to find the factors of x 3 − 6x 2 + 11x − 6 4. Determine the factors of x 3 + 7x 2 + 14x + 8 and hence solve the cubic equation x 3 + 7x 2 + 14x + 8 = 0 5. Determine the value of ‘a’ if (x + 2) is a factor of (x 3 − ax 2 + 7x + 10) 6. com/cw/bird Chapter 2 Partial fractions Why it is important to understand: Partial fractions The algebraic technique of resolving a complicated fraction into partial fractions is often needed by electrical and mechanical engineers for not only determining certain integrals in calculus, but for determining inverse Laplace transforms, and for analysing linear differential equations with resonant circuits and feedback control systems.

E. 6903 . . 7183) are called hyperbolic, Napierian or natural logarithms, and loge is usually abbreviated to ln. The following values may be checked by using a calculator: Problem 4. e. 2947 . 0338 . . 7037 . . More on Napierian logarithms is explained in Chapter 4. Here are some worked problems to help understanding of logarithms. using a calculator) Problem 5. e. from which, x = 1 Problem 1. Evaluate log3 9 Hence, Let x = log3 9 then 3 x = 9 e x = e1 from the definition of a logarithm, ln e = 1 (which may be checked using a calculator) Section A Logarithms Section A 24 Higher Engineering Mathematics Problem 6.