Questions and Answers #12 Polynomial Arithmetic

Questions and Answers Sheet 12 Polynomial Arithmetic Question #1 Subtract polynomials f(x)-g(x)=? Answer: Assuming 𝑓(𝑥) = 3𝑥 2 − 10𝑥 + 5 𝑔(𝑥) = 6𝑥 2 + 8𝑥 − 12 When we perform f(x) - g(x) = (3𝑥 2 − 10𝑥 + 5) − (6𝑥 2 + 8𝑥 − 12) When you have polynomials having x terms with different powers, just group the same terms and perform the arithmetic operation. if 𝑓(𝑥) = 𝑎𝑥 2 + 𝑐𝑥 + 𝑒 𝑔(𝑥) = 𝑏𝑥 2 + 𝑑𝑥 + 𝑓 then 𝑓(𝑥) − 𝑔(𝑥) = (𝑎 − 𝑏)𝑥 2 + (𝑐 − 𝑑)𝑥 + (𝑒 − 𝑓) This can be done on polynomials of different exponents, just remember to perform operations on same variable terms. Therefore, 𝑓(𝑥) − 𝑔(𝑥) = (3 − 6)𝑥 2 + ((−10) − 8)𝑥 + (5 − (−12)) = −3𝑥 2 − 18𝑥 2 + 17 Question #2 The terms in the polynomial −4𝑦 5 + 7𝑦 2 − 3𝑦 − 2. Answer: A term in a polynomial is separated by arithmetic operations. A term can be a constant value, any arbitrary variable like a, multiplication of constant value and variable or division of a constant value and variable provided that constant value should be nonzero if it's present in denominator. In the polynomial −4𝑦 5 + 7𝑦 2 − 3𝑦 − 2, the terms are: −4𝑦 5 ; The constant term is -4 and in multiplication variable is y raised to power 5. 7𝑦 2 : The constant term is 7 and in multiplication variable is y raised to power 2. -3y : The constant term is -3 and in multiplication variable is y raised to power 1. -2 : The constant term is -2. Hence, the terms of the polynomial −4𝑦 5 + 7𝑦 2 − 3𝑦 − 2 are − 4𝑦 5 , 7𝑦 2 , −3𝑦 and − 2. Question #3 To calculate: Polynomial equation with real coefficients that has roots 3, 1-i. Answer: Formula used: (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Calculation: If the polynomial has real coefficients, then it's imaginary roots occur in conjugate pairs. So, a polynomial with the given root 1-i must have another root as 1+i . Since each root of the equation corresponds to a factor of the polynomial, also, the roots indicate zeros of that polynomial, thus, the polynomial equation is written as, (𝑥 − 3)[𝑥 − (1 − 𝑖)][𝑥 − (1 + 𝑖)] = 0 (𝑥 − 3)(𝑥 − 1 + 𝑖)(𝑥 − 1 − 𝑖) = 0 Further use arithmetic rule, (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Here, a=x-1, b=i Now, the polynomial equation is, (𝑥 − 3)[(𝑥 − 1)2 − (𝑖)2 ] = 0 Use arithmetic rule. (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 and 𝑖 2 = −1. Now, the polynomial equation is: (𝑥 − 3)(𝑥 2 − 2𝑥 + 1 + 1) = 0 (𝑥 − 3)(𝑥 2 − 2𝑥 + 2) = 0 𝑥 3 − 3𝑥 2 − 2𝑥 2 + 6𝑥 + 2𝑥 − 6 = 0 𝑥 3 − 5𝑥 2 + 8𝑥 − 6 = 0 Hence, the polynomial equation of given roots 3, 1 − 𝑖 is 𝑥 3 − 5𝑥 2 + 8𝑥 − 6 = 0. Question #4 Find a polynomial equation with real coefficients that has the given roots. -1, 4-2i Answer: (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Calculation: If the polynomial has real coefficients, then it's imaginary roots occur in conjugate pairs. So, a polynomial with the given root 4-2i must have another root as 4+2i . Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial. Hence, below mentioned equation is written as. [𝑥 − (−1)][𝑥 − (4 − 2𝑖)][𝑥 − (4 + 2𝑖)] = 0 (𝑥 + 1)(𝑥 − 4 + 2𝑖)(𝑥 − 4 − 2𝑖) = 0 Further use arithmetic rule. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Here, a=x-4, b=2i Now, the polynomial equation is, (𝑥 + 1)[(𝑥 − 4)2 − (2𝑖)2 ] = 0 Use arithmetic rule. (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 And 𝑖 2 = −1. Now, the polynomial equation is, (𝑥 + 1)(𝑥 2 − 2𝑥 + 16 + 4) = 0 (𝑥 + 1)(𝑥 2 − 8𝑥 + 20) = 0 (𝑥 3 + 𝑥 2 − 8𝑥 2 − 8𝑥 + 20𝑥 + 20) = 0 𝑥 3 − 7𝑥 2 + 12𝑥 + 20 = 0 Hence, the polynomial equation of given roots −1,4 − 2𝑖 is 𝑥 3 − 7𝑥 2 + 12𝑥 + 20 = 0. Question #5 Find a polynomial equation with real coefficients that has the given roots. 1,2-3i Answer: (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Calculation: If the polynomial has real coefficients, then it's imaginary roots occur in conjugate pairs. So, a polynomial with the given root 2-3i must have another root as 2+3i . Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial. Hence, below mentioned equation is written as. (x-1)[x-(2-3i)][x-(2+3i)]=0 (x-1)(x-2+3i)(x-2-3i)=0 Further use arithmetic rule. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Here, a=x-2, b=3i Now, the polynomial equation is, (𝑥 − 1)[(𝑥 − 2)2 − (3𝑖)2 ] = 0 Use arithmetic rule. (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 And 𝑖 2 = −1. Now, the polynomial equation is, (𝑥 − 1)(𝑥 2 − 4𝑥 + 4 + 9) = 0 (𝑥 − 1)(𝑥 2 − 4𝑥 + 13) = 0 (𝑥 3 − 𝑥 2 − 4𝑥 2 + 4𝑥 + 13𝑥 − 13) = 0 𝑥 3 − 5𝑥 2 + 17𝑥 − 13 = 0 Hence, the polynomial equation of given roots 1,2 − 3𝑖 is 𝑥 3 − 5𝑥 2 + 17𝑥 − 13 = 0. Question #6 To calculate: Polynomial equation with real coefficients that has roots 5, 4-3i . Answer: Formula Used: (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Calculation: If the polynomial has real coefficients, then it's imaginary roots occur in conjugate pairs. So, a polynomial with the given root 4-3i must have another root as 4+3i . Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial. Hence, below mentioned equation is written as. (x-5)[x-(4-3i)][x-(4+3i)]=0 (x-5)(x-4+3i)(x-4-3i)=0 Further use arithmetic rule. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Here, a=x-4, b=3i Now, the polynomial equation is, (𝑥 − 5)[(𝑥 − 4)2 − (3𝑖)2 ] = 0 Use arithmetic rule. (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 And 𝑖 2 = −1. Now, the polynomial equation is, (𝑥 − 5)(𝑥 2 − 8𝑥 + 16 + 9) = 0 (𝑥 − 5)(𝑥 2 − 8𝑥 + 25) = 0 𝑥 3 − 5𝑥 2 − 8𝑥 2 + 40𝑥 + 25𝑥 − 125 = 0 𝑥 3 − 13𝑥 2 + 65𝑥 − 125 = 0 Hence, the polynomial equation of given roots 5,4 − 3𝑖 is 𝑥 3 − 13𝑥 2 + 65𝑥 − 125 = 0. Question #7 To calculate: Polynomial equation with real coefficients that has roots 3i, 3-i Answer: Formula Used: (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Calculation: If the polynomial has real coefficients, then it's imaginary roots occur in conjugate pairs. So, a polynomial with the given root 3i,3-i must have another root as -3i, 3+i. Since each root of the equation corresponds to a factor of the polynomial also the roots indicate zeros of that polynomial. Hence, below mentioned equation is written as. (x-3i)[x-(-3i)][x-(3-i)][x-(3+i)]=0 (x-3i)(x+3i)(x-3+i)(x-3-i)=0 Further use arithmetic rule. (𝑎 + 𝑏)(𝑎 − 𝑏) = 𝑎2 − 𝑏 2 Now, the polynomial equation is, [𝑥 2 − (3𝑖)2 ][(𝑥 − 3)2 − (𝑖)2 ] = 0 Use arithmetic rule. (𝑎 − 𝑏)2 = 𝑎2 − 2𝑎𝑏 + 𝑏 2 And 𝑖 2 = −1. Now, [𝑥 2 + 9][𝑥 2 − 6𝑥 + 9 + 1] = 0 (𝑥 2 + 9)(𝑥 2 − 6𝑥 + 10) = 0 (𝑥 4 + 9𝑥 2 − 6𝑥 3 − 54𝑥 + 10𝑥 2 + 90) = 0 𝑥 4 − 6𝑥 3 + 19𝑥 2 − 54𝑥 + 90 = 0 Hence, the polynomial equation of given roots 3𝑖, 3 − 𝑖 is 𝑥 4 − 6𝑥 3 + 19𝑥 2 − 54𝑥 + 90 = 0. Question #8 Expand Q to prove that the polynomials P and Q are the same. 𝑃(𝑥) = 3𝑥 4 − 5𝑥 3 + 𝑥 2 − 3𝑥 + 5 𝑄(𝑥) = (((3𝑥 − 5)𝑥 + 1)𝑥 − 3) 𝑥 + 5 Try to evaluate P(2) and Q(2) in your head, using the forms given. Which is easier? Now write the polynomial 𝑅(𝑥) = 𝑥 5 − 2𝑥 4 + 3𝑥 3 − 2𝑥 2 + 3𝑥 + 4 in "nested" form, like the polynomial Q. Use the nested form to find R(3) in your head. Do you see how calculating with the nested form follows the same arithmetic steps as calculating the value of a polynomial using synthetic division? Answer: Given: 𝑃(𝑥) = 3𝑥 4 − 5𝑥 3 + 𝑥 2 − 3𝑥 + 5 𝑄(𝑥) = (((3𝑥 − 5)𝑥 + 1)𝑥 − 3) 𝑥 + 5 𝑅(𝑥) = 𝑥 5 − 2𝑥 4 + 3𝑥 3 − 2𝑥 2 + 3𝑥 + 4 To find P(2): 𝑃(𝑥) = 3𝑥 4 − 5𝑥 3 + 𝑥 2 − 3𝑥 + 5 𝑃(2) = 3(2)4 − 5(2)3 + (2)2 − 3(2) + 5 =3(16)-5(8)+4-6+5 =48-40+4-6+5 =57-46 =11 To find Q(2): Q(x)=(((3x-5)x+1)x-3)x+5 Q(2)=(((3(2)-5)2+1)2-3)2+5 =(((1)2+1)2-3)2+5 (6-3)2+5 3(2)+5 6+5 =11 Now we have to write R(x) in nested form. 𝑅(𝑥) = 𝑥 5 − 2𝑥 4 + 3𝑥 3 − 2𝑥 2 + 3𝑥 + 4 Now we have to factor out x. 𝑅(𝑥) = (𝑥 4 − 2𝑥 3 + 3𝑥 2 − 2𝑥 + 3)𝑥 + 4 = ((𝑥 3 − 2𝑥 2 + 3𝑥 − 2)𝑥 + 3)𝑥 + 4 = (((𝑥 2 − 2𝑥 + 3)𝑥 − 2)𝑥 + 3) 𝑥 + 4 = ((((𝑥 − 2)𝑥 + 3)𝑥 − 2) 𝑥 + 3) 𝑥 + 4 𝑅(𝑥) = ((((𝑥 − 2)𝑥 + 3)𝑥 − 2) 𝑥 + 3) 𝑥 + 4 𝑅(3) = ((((3 − 2)3 + 3)3 − 2) 3 + 3) 3 + 4 = ((((1)3 + 3)3 − 2) 3 + 3) 3 + 4 = (((6)3 − 2)3 + 3) 3 + 4 = ((18 − 2)3 + 3)3 + 4 ((16)3 + 3)3 + 4 = (48 + 3)3 + 4 = (51)3 + 4 = 153 + 4 = 157 𝑅(3) = 157 Hence, The polynomial P and Q are the same by expanding Q is proved. Question #9 𝑝 𝑟 𝑞 𝑠 a) The missing terms for the given polynomial p,q,r, and s such that 𝑞 ≠ 0 and 𝑠 ≠ 0, ⋅ =?. 𝑝 𝑟 𝑞 𝑠 b) The missing terms given polynomial p,q,r, and s such that 𝑞 ≠ 0, 𝑟 ≠ 0 and 𝑠 ≠ 0, ÷ =?. Answer: a) Concept used: Polynomials can be multiplied by applying the distributive property of multiplication over addition and then combining the like terms. The resulting expression is a polynomial of degree equal to sum of the degrees of the two polynomials. Calculation: 𝑝 𝑟 Consider two rational functions 𝑓(𝑥) = and 𝑔(𝑥) = , such that the polynomials q and s are not zero, 𝑞 𝑠 then the product 𝑓(𝑥) ⋅ 𝑔(𝑥) will be a rational function defined by 𝑝 𝑟 𝑝𝑟 𝑞 𝑠 𝑞𝑠 such that 𝑞 ≠ 0 and 𝑠 ≠ 0, ⋅ = 𝑝 𝑞 𝑟 ⋅ . Thus, for polynomials p,q,r, and s 𝑠 . 𝑝 𝑟 𝑝𝑟 𝑞 𝑠 𝑞𝑠 Conclusion: The polynomials p,q,r and s such that 𝑞 ≠ 0 and 𝑠 ≠ 0, ⋅ = .