Questions and Answers #49 Polynomial Factorization

Questions and Answers Sheet 49 Polynomial Factorization Question #1 Solve the following quadratic equations by factorization method: 𝑥 + 3 1 − 𝑥 17 − = . 𝑥−2 𝑥 4 Answer: Step 1 The quadratic equation is given by 𝑥 + 3 1 − 𝑥 17 − = . 𝑥−2 𝑥 4 To evaluate: The solution of the quadratic equation by factorization method. Step 2 The quadratic equation can be further simplified and solved by factorization method as following : 𝑥 + 3 1 − 𝑥 17 − = . 𝑥−2 𝑥 4 (𝑥+3)𝑥−(1−𝑥)(𝑥−2) 17 ⇒ = (𝑥−2)𝑥 4 ⇒ ⇒ (𝑥 2 +3𝑥)−(−𝑥 2 +3𝑥−2) (𝑥 2 −2𝑥) 𝑥 2 +3𝑥+𝑥 2 −3𝑥+2 (𝑥 2 −2𝑥) 2 = = 17 4 17 4 ⇒ 4(2𝑥 + 2) = 17(𝑥 2 − 2𝑥) ⇒ 8𝑥 2 + 8 = 17𝑥 2 − 34𝑥 ⇒ 9𝑥 2 − 34𝑥 − 8 = 0 ⇒ 9𝑥 2 − 36𝑥 + 2𝑥 − 8 = 0 ⇒ 9𝑥(𝑥 − 4) + 2(𝑥 − 4) = 0 ⇒ (9𝑥 + 2)(𝑥 − 4) = 0 ⇒ (9𝑥 + 2) = 0 or (x-4)=0 ⇒𝑥=− 2 9 or x=4 Step 3 Hence, the solutions of the quadratic equation are: 𝑥 = − 2 9 or x=4 Question #2 All the real zeros of the given polynomial are integers. Find the zeros. (Enter your answers as a commaseparated list. Enter all answers including repetitions.) 𝑃(𝑥) = 𝑥 3 + 5𝑥 2 − 𝑥 − 5 x= Write the polynomial in factored form. P(x) = Answer: Step 1 Analysis Given: 𝑃(𝑥) = 𝑥 3 + 5𝑥 2 − 𝑥 − 5 Step 2 Factorisation of P(x). Factoring P(x), 𝑥 3 + 5𝑥 2 − 𝑥 − 5 = 𝑥 2 (𝑥 + 5) − (𝑥 + 5) = (𝑥 2 − 1)(𝑥 + 5) =(x-1)(x+1)(x+5) The factorisation of P(x) is (x-1)(x+1)(x+5) Step 3 Zeroes of P(x). Equating P(x)=0, (x-1)(x+1)(x+5)=0 Therefore, the zeroes of P(x) are x=1, x=-1 and x=-5 Question #3 What is the Greatest Common Factor (GCF) in this trinomial? 12𝑥 2 − 32𝑥 − 56 Answer: Step 1 Given 12𝑥 2 − 32𝑥 − 56 We have to find the greatest common factor in the trinomial. Step 2 We have 12𝑥 2 − 32𝑥 − 56 ⇒ 4 × 3 × 𝑥 2 − 4 × 8 × 𝑥 − 4 × 14 (done factorization) ⇒ 4 × (3 × 𝑥 2 − 8 × 𝑥 − 14) (taken 4 as common) ⇒ 4 × (3𝑥 2 − 8𝑥 − 14) ⇒ 4(3𝑥 2 − 8𝑥 − 14) So, The greatest common factor is 4 Answer Question #4 Solving the following quadratic equation by factoring. 𝑥 2 = 4𝑥 + 5 Answer: Step 1 We have to solve the quadratic equation by factorization and we also have to check the answer where quadratic equation is: 𝑥 2 = 4𝑥 + 5 Rewriting the quadratic equation, 𝑥 2 = 4𝑥 + 5 𝑥 2 − 4𝑥 − 5 = 0 Now solving by factorization of middle term splitting method, we get 𝑥 2 − 4𝑥 − 5 = 0 𝑥 2 − 5𝑥 + 𝑥 − 5 = 0 𝑥(𝑥 − 5) + 1(𝑥 − 5) = 0 (x-5)(x+1)=0 Either, x-5=0 x=5 or, x+1=0 x=-1 Hence, solutions of the quadratic equation are x=-1, 5. Step 2 Checking the answer: Putting x=-1 into the quadratic equation, we get 𝑥 2 − 4𝑥 − 5 = 0 (−1)2 − 4(−1) − 5 = 0 1+4-5=0 5-5=0 0=0 Which is satisfying the given equation. Hence, x=-1 is the correct answer. Putting x=5 into the quadratic equation, we get 52 − 4 × 5 − 5 = 0 25-20-5=0 5-5=0 0=0 Which is satisfying the given equation. Hence, x=5 is the correct answer. Both solution are correct. Question #5 Factor trinomial, or state that the trinomial is ': 3𝑥 2 + 4𝑥𝑦 + 𝑦 2 Answer: Step 1 We have to factor the trinomial or state that the trinomial is ': 3𝑥 2 + 4𝑥𝑦 + 𝑦 2 Factorization is the process of taking common and that common parts are known as factor of that expression. Factorizing the trinomial by middle term splitting method, In middle term splitting method first we multiply first and last term constant coefficients then we do ' factor of that multiplied part and after adding or subtracting we substitute the middle term. Here first and last term constant coefficient are 3 and 1 respectively Therefore, 3 × 1 So for middle term 4, 3+1=4 Step 2 Now factorizing the trinomial by using middle term splitting method, 3𝑥 2 + 4𝑥𝑦 + 𝑦 2 = 3𝑥 2 + 3𝑥𝑦 + 𝑥𝑦 + 𝑦 2 =3x(x+y)+y(x+y) =(x+y)(3x+y) Hence, factorization of the trinomial is (x+y)(3x+y). Note: If it was not factorable then it could be '. Question #6 Which of the polynomials below is a factor of the polynomial? 12𝑥 4 − 8𝑥 2 − 9𝑥 2 + 6 A. 3𝑥 2 + 2 B. 4𝑥 4 − 3 C. 𝑥 2 + 2 D. 4𝑥 2 − 3 E. The polynomal is ' F. None of these Answer: Step 1 We have to factor Step 2 12𝑥 4 − 8𝑥 2 − 9𝑥 2 += 12𝑥 4 − 17𝑥 2 + 6 Let𝑥 2 = 𝑘 so 𝑥 4 = 𝑘 2 That means 12𝑘 2 − 17𝑘+= 12𝑘 2 − 8𝑘 − 9𝑘 + 6 = 4k(3k-2)-3(3k-2) = (4k-3)(3k-2) = (4𝑥 2 − 3)(3𝑥 2 − 2) Ans: Factor is 4𝑥 2 − 3 Question #7 Find all rational zeros of the polynomial, and then find the irrational zeros, if any. Whenever appropriate, use the Rational Zeros Theorem, the Upper and Lower Bounds Theorem, Descartes' Rule of Signs, the Quadratic Formula, or other factoring techniques. 𝑃(𝑥) = 2𝑥 4 + 3𝑥 3 − 4𝑥 2 − 3𝑥 + 2 Answer: Step 1 Zeros of the polynomial are those values which return zero value for the function when put in place for independent variable. Zeroes of a polynomial are values for which function f(x) returns value 0. Highest power of polynomial is the degree of the polynomial. Degree of polynomial is the highest power that the polynomial has. Leading coefficient is the coefficient of the term in polynomial which has highest power. X-intercept is the point where graph of polynomial function touches/crosses x-axis, i.e y=0. Y-intercept is the point where graph of polynomial function touches/crosses y-axis-axis, i.e x=0 Step 2 We can simplify the given polynomial as 𝑃(𝑥) = 2𝑥 4 + 3𝑥 3 − 4𝑥 2 − 3𝑥 + 2 By trial and error we can find that P(1)=0, hence (x-1) is factor of the polynomial. 𝑃(𝑥) = (𝑥 − 1)(2𝑥 3 + 5𝑥 2 + 𝑥 − 2) Again by trial error we can find, P(-1)=0, hence (x+1) is also factor of above polynomial 𝑃(𝑥) = (𝑥 − 1)(𝑥 + 1)(2𝑥 2 + 3𝑥 − 2) Solving the quadratic, we get 𝑃(𝑥) = (𝑥 − 1)(𝑥 + 1)(2𝑥 − 1)(𝑥 + 2) Hence the rational zeros of the polynomial are x=1, -1, 1/2, -2. Polynomial has no irrational zeros. Question #8 Find all the zeros of the polynomial: 𝑓(𝑥) = 𝑥 3 + 4𝑥 2 − 41𝑥 + 36 ... if (x-4) is a factor of this polynomial. Answer: Step 1 Given, 𝑓(𝑥) = 𝑥 3 + 4𝑥 2 − 41𝑥 + 36 (x-4) is a factor of this polynomial We have to find the all the zeros of the polynomial. Step 2 We have 𝑓(𝑥) = 𝑥 3 + 4𝑥 2 − 41𝑥 + 36 (x-4) is a factor of this polynomial So, Factorizing 𝑓(𝑥) = 𝑥 3 + 4𝑥 2 − 41𝑥 + 36 further ⇒ 𝑥 3 + 4𝑥 2 − 41𝑥 + 36 ⇒ 𝑥 3 − 4𝑥 2 + 8𝑥 2 − 32𝑥 − 9𝑥 + 36 ⇒ 𝑥 2 (𝑥 − 4) + 8𝑥(𝑥 − 4) − 9(𝑥 − 4) ⇒ (𝑥 − 4)(𝑥 2 + 8𝑥 − 9) ⇒ (𝑥 − 4)(𝑥 2 + 9𝑥 − 𝑥 − 9) ⇒ (𝑥 − 4)(𝑥(𝑥 + 9) − 1(𝑥 + 9)) ⇒ (𝑥 − 4)(𝑥 + 9)(𝑥 − 1) Step 3 Now, For finding the zeros of the f(x) Then, ⇒ 𝑓(𝑥) = 0 ⇒ (𝑥 − 4)(𝑥 − 1)(𝑥 + 9) = 0 ⇒ 𝑥 − 4 = 0 or x-1=0 or x+9=0 ⇒ 𝑥 = 4 or x=1 or x=-9 So, The zhe zeros the of f(x) is x = 4, 1 and -9 Answer Question #9 Use the polynomials below to find the sum of the given polynomials. Write the type of polynomial. 𝑓(𝑥) = 𝑥 3 + 2𝑥 2 − 7𝑥 − 8 g(x)=-4x-7 ℎ(𝑥) = 4𝑥 2 − 𝑥 − 15 𝑘(𝑥) = −𝑥 2 + 7𝑥 + 4 a) h(x)+k(x) b) g(x)+f(x) c) f(x)+k(x) Answer: Step 1 Given: 𝑓(𝑥) = 𝑥 3 + 2𝑥 2 − 7𝑥 − 8, 𝑔(𝑥) = −4𝑥 − 7, ℎ(𝑥) = 4𝑥 2 − 𝑥 − 15, 𝑘(𝑥) = −𝑥 2 + 7𝑥 + 4 For finding (h(x)+k(x)) function, we add both functions Step 2 So, ℎ(𝑥) + 𝑘(𝑥) = (4𝑥 2 − 𝑥 − 15) + (−𝑥 2 + 7𝑥 + 4) = (4𝑥 2 − 𝑥 2 ) + (−𝑥 + 7𝑥) + (−15 + 4) = 3𝑥 2 + 6𝑥 − 11 = 3𝑥 2 + 6𝑥 − 11 … (1) Now, by equation(1) we can see that (h(x)+k(x)) is quadratic function Similarly, 𝑔(𝑥) + 𝑓(𝑥) = (−4𝑥 − 7) + (𝑥 3 + 2𝑥 2 − 7𝑥 − 8) = 𝑥 3 + 2𝑥 2 + (−4𝑥 − 7𝑥) + (−7 − 8) = 𝑥 3 + 2𝑥 2 − 11𝑥 − 15 … (2) Now, by equation(2) we can see that (g(x)+f(x)) is cubic function Step 3 Similarly, 𝑓(𝑥) + 𝑘(𝑥) = (𝑥 3 + 2𝑥 2 − 7𝑥 − 8) + (−𝑥 2 + 7𝑥 + 4) = 𝑥 3 + (2𝑥 2 − 𝑥 2 ) + (−7𝑥 + 7𝑥) + (−8 + 4) = 𝑥3 + 𝑥2 + 0 − 4 = 𝑥3 + 𝑥2 − 4 Now, by equation(3) we can see that (f(x)+k(x)) is also cubic function Hence, given functions can be find as above. Question #10 A polynomial is given. 3𝑥 5 + 2𝑥 2 What is the degree of this polynomial? A) 2 B) 3 C) 5 D) 7 Answer: Step 1 Given, A polynomial is 3𝑥 5 + 2𝑥 2 . Step 2 Now, Since, the degree of a polynomial in one variable is the largest exponent in the polynomial. So, In the polynomial: 3𝑥 5 + 2𝑥 2 Degree = 5 ∴ Option (c) 5 is correct.