Polynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if


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1 1. Polynomials 1.1. Definitions A polynomial in x is an expression obtained by taking powers of x, multiplying them by constants, and adding them. It can be written in the form c 0 x n + c 1 x n 1 + c 2 x n c n 1 x + c n where n is an integer 0, and c 0, c 1,..., c n are constants. For example 2x 6 + 5x 5 + x x2 6x 10 Note that x and 4x are not polynomials. x 2 The constants c 0,..., c n are called the coefficients. The constant term is c n. The leading term is the term involving the highest power of x, here c 0 x n. The degree is the power of x in the leading term. A degree 0 polynomial is just a constant, e.g., 2 is (a) constant A degree 1 polynomial is called linear, e.g., 3x + 2 is linear A degree 2 polynomial is called quadratic, e.g., x 2 + 2x + 1 is quadratic A degree 3 polynomial is called cubic, e.g., y 3 + 7y 2 is a cubic in y Operations Polynomials can be added or subtracted simply by adding or subtracting the corresponding terms, e.g., if then f(x) = x 2 3, g(x) = 4x 3 + 3x 2 2x + 6 f(x) + g(x) = 4x 3 + 4x 2 2x + 3, f(x) g(x) = 4x 3 2x 2 + 2x 9 To multiply two polynomials, every term of the first polynomial must be multiplied by every term of the second. Example 1.1. f(x)g(x) =(x 2 3)(4x 3 + 3x 2 2x + 6) =x 2 (4x 3 + 3x 2 2x + 6) 3(4x 3 + 3x 2 2x + 6) =4x 5 + 3x 4 2x 3 + 6x 2 12x 3 9x 2 + 6x 18 =4x 5 + 3x 4 14x 3 3x 2 + 6x 18 1
2 1.3. Polynomial division Similar to long division for ordinary numbers. To divide 2x x 2 3x + 1 by x + 3, we compute 2x 2 +4x 15 x + 3 2x x 2 3x +1 2x 3 + 6x 2 4x 2 3x 4x 2 +12x 15x +1 15x The term 2x 3 must come from multiplying x+3 by 2x 2, so put 2x 2 on the top line, multiply x + 3 by 2x 2 and subtract from the line above. Now bring down the 3x. Multiply x + 3 by 4x and subtract. Bring down the 1. Multiply x + 3 by 15 and subtract. This completes the process. The quotient is 2x 2 + 4x 15 and the remainder is 46. polynomial = divisor quotient + remainder 2x x 2 3x + 1 = (x + 3)(2x 2 + 4x 15) + 46 Note. If the remainder is zero, we say that the polynomial is exactly divisible by the divisor, or that the divisor is a factor of the polynomial. Example 1.2. Divide x 4 2x 2 + 3x 6 by x 2 4x + 3. x 2 +4x +11 x 2 4x + 3 x 4 2x 2 +3x 6 x 4 4x 3 +3x 2 4x 3 5x 2 +3x 4x 3 16x 2 +12x 11x 2 9x 6 11x 2 44x x 39 The quotient is x 2 + 4x The remainder is 35x 39. polynomial = divisor quotient + remainder x 4 2x 2 + 3x 6 = (x 2 4x + 3)(x 2 + 4x + 11) + 35x 39 Note. It is always true that degree of remainder < degree of divisor. 2
3 1.4. Remainder Theorem Theorem 1.3. When a polynomial f(x) is divided by x a, then the remainder is f(a). Example 1.4. The remainder when f(x) = 2x 5 3x is divided by x + 2 is f( 2) = ( 2) 5 3( 2) = = 43. Check this yourself by doing the long division. Proof. If the quotient is q(x) and the remainder is r, then f(x) = (x a).q(x) + r Substituting x = a, gives f(a) = (a a) q(a) + r = r Factor Theorem Theorem 1.5. If x a is a factor of a polynomial f(x), then f(a) = 0. Conversely, if f(a) = 0, then x a is a factor of f(x). Example 1.6. Is x 1 is a factor of f(x) = x 5 1? Since f(1) = 0, the Factor Theorem tells us it must be. In fact x 5 1 = (x 1)(x 4 + x 3 + x 2 + x + 1). Proof. If x a is a factor of f(x) then f(x) = (x a) g(x) for some polynomial g(x). Therefore f(a) = (a a) g(a) = 0. If f(a) = 0 then by the Remainder Theorem, the remainder on dividing f(x) by x a is 0. Therefore f(x) = (x a) q(x) where q(x) is the quotient Factorizing polynomials It is often useful to write a polynomial as a product of polynomials of lower degree. For example f(x) = x 3 2x 2 x + 2 can be factorized as (x 2 1)(x 2) or as (x 1)(x + 1)(x 2). Factorizing helps find the roots since, if a product of numbers is zero, then one of them must be zero. Therefore if f(x) = 0 then x 1 = 0, or x + 1 = 0 or x 2 = 0. Therefore x = 1, 1 or 2. Some wellknown factorizations: (i) Difference of squares: x 2 a 2 = (x a)(x + a) for example, x 2 9 = (x + 3)(x 3). (ii) Perfect square: x 2 ± 2ax + a 2 = (x ± a) 2 for example x 2 10x + 25 = (x 5) 2. (iii) If there is no constant term, then x is a common factor. For example, x 2 + 4x = x(x + 4). 3
4 1.7. Factorizing by trial and error Given a quadratic x 2 + bx + c, we try to write it as x 2 + bx + c = (x + r)(x + s) for some r, s. Comparing coefficients, we need to find two numbers r and s with r + s = b and rs = c. If r and s are integers, they must be factors of c. Sometimes they can be found by trial and error. Example 1.7. x 2 + 7x Need rs = 10. You can write 10 = 1 10 = 2 5. The second one works since = 7. Therefore x 2 + 7x + 10 = (x + 2)(x + 5). Example 1.8. x 2 + 3x 18. Need rs = 18, so one of the numbers r and s must be positive, the other negative. Now 18 = 1 18 = 2 9 = 3 6. In fact r = 6, s = 3 works, since r + s = 3. Therefore x 2 + 3x 18 = (x + 6)(x 3). To factorize ax 2 + bx + c with a 1 by inspection is harder, but it can sometimes be done. Example x 2 + 7x + 6. Try to write it as (2x + r)(x + s). Need rs = 6 and r + 2s = 7. Take r = 3 and s = 2, so 2x 2 + 7x + 6 = (2x + 3)(x + 2) Using the Factor Theorem If f(a) = 0 then x a is a factor of f(x). If f(x) = ax 2 + bx + c has roots x = x 1, x 2, then f(x) = a(x x 1 )(x x 2 ). Example From the standard formula, x 2 + 6x 72 has roots x = 6 ± 324 giving 2 x = 6, 12. Therefore x 2 + 6x 72 = (x 6)(x + 12). It can be difficult to factorize polynomials of degree 3. If you happen to know a root, say f(a) = 0, then x a is a factor, so f(x) = (x a) q(x). Use polynomial division to find q(x), and try to factorize it Worked examples Example Find the quotient and remainder for (x 3 + 5x 2 + 4x 17) (x 4). Example Find the quotient and remainder for (x 3 4x 2 + x + 2) (x 2 3). Example When x 3 4x 2 + 5x + a is divided by x 3, the remainder is 7. Find a. Example Factorize the following polynomials: x 2 + 7x + 12, x 2 + 8x, x 3 5x 2 50x, 2x 2 5x + 3, x 3 + 3x 2 4x 12. Example The polynomial x 3 + 8x 2 + kx + 10 has a factor x + 2. (a) Find the value of k. (b) Factorize the polynomial. 4
5 Example Given that f(x) = x 3 4x 2 x+4, find f(1) and f(2), and hence factorize f(x) into a product of three linear factors. Example Express 6x x x + 3 in the form (x + 1)(ax 2 + bx + c), stating the values of a, b and c, and hence solve the equation 6x x x + 3 = 0. 5
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