Factorizing a Polynomial. What is Quadratic Equation? What is Polynomial Factorization? Factorizing by grouping like terms. Factorizing by identities

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1 Chapter 3 Quadratic Equation and Polynomial Factorization What is Quadratic Equation? Quadratic equation ( 二次方程 ) is a type of polynomials. It is an equation that the left-hand-side is a polynomial with degree, and usually unknown, and the right-hand-side is a zero. A general epression of quadratic equation is: a + b + c = 0 Where a, b and c are some constants, and is the variable. What is Polynomial Factorization? Polynomial Factorization ( 多項式因式分解 ) is the action that change a polynomial into products of smaller polynomials. For eample: can be factorized into: ( 6) ( + ) ( + ) Here, ( + 6), ( + ) and ( + ) are called factors ( 因子 ) of Factorization can be considered as the reverse action of multiplication. Factorizing a Polynomial Factorizing by grouping like terms We know that = ( 6) ( + ) ( + ) now. But how is it found? First, review some basic facts: n = a (b + c) = ab + ac n We ll factorize by these facts = ( 6) ( 6) 3 ( )( 6) = + = + Now we ve successfully factorized into ( 6) ( 3 + ) by using the method Grouping like terms. Factorizing by identities But clearly we haven t finished. The term ( 3 + ) can be further factored. How do we factorize it? Grouping terms seems not working! So, instead of grouping terms, we may also need some help from identities. From chapter, we learnt that:. (a + b) = a + ab + b. (a b) = a ab + b 3. (a + b)(a b) = a b For eample, when we see the epression 4, we knew that it could be factorized into ( + )( ), because of identity #3. Here we present two more identities: 5. a 3 + b 3 = (a + b)(a ab + b ) 6. a 3 b 3 = (a b)(a + ab + b ) Thus, by identity #5, we can reduce ( 3 + ) into ( + ) ( + ), and we have fully factorized

2 Application of Factorization One big use of polynomial factorization is to find roots ( 根 ) (solutions) of a polynomial. For eample, we need to solve = 0. It is hard to do so. However, we knew that if some numbers are multiplied and the result is zero, then one of them must be zero. (i.e., if a b c d e f = 0, then a = 0 or b = 0 or c = 0 or d = 0 or e = 0 or f = 0 or ) This also applies for polynomials. Thus, if = 0 Then also ( 6) ( + ) ( + ) = 0 That means 6 = 0 or + = 0 or + = 0 Solving the first equations gives = 6 or. The last equation gives no real roots ( 實根 ) of, and we ll eplain why in the later section. Another reason for factorizing a polynomial is to reduce computation time. For eample, calculating the polynomial needs 8 multiplications and 3 additions (including subtraction). In contrast, the factorized form ( 6) ( + ) ( + ) only needs 3 multiplications and 4 additions. Since time needed for doing multiplications is much more than additions, computation time is reduced. Last but not least, factorizing a polynomial is also useful in simplifying an epression. For eample, is a very tedious epression. But if we factorize it, it ( )( )( 3) becomes. The factors ( ) and ( 3) can be cancelled out, and the ( )( )( ) ( ) 5( + 8) remaining part is just simply. Solving a Quadratic Equation In general, solving a quadratic equation a + b + c = 0 can be done in several ways. We will use = 0 in our eample. Cross method ( 十字相乘法 ) The cross method is to solve a quadratic equation by factorizing. We may assume that = ( α) ( β). Multiplying the right-hand-side, we get = (α + β) + αβ. α + β = 4 By comparing coefficients, we get. However, method of substitution do not αβ = 40 work here, because we don t know how to solve 4α α = 40 yet. But, we may use a low-level method to find out α and β: by guessing. Of course we shouldn t guess randomly. From the simultaneous equation above, we knew that αβ = 40, that means, α and β are factors of 40. Therefore, α and β may be,, 4, 5, 8, 0, 0, 40, -, -, -4, -5, -8, -0, -0 or -40. When α = β =

3 Net step is to select to right solution, by the condition α + β = 4. (, 40) is not a solution, because + 40 = 4 4. (, 0) is not either. (4, 0) is the solution. As soon as we found out the correct one, we can stop. Therefore, = ( 4) ( 0), and thus = 4 or 0. Completing the square ( 配方 ) Trial and error is easy to follow, but it is etremely inefficient. Imagine if we have a long list to test, such as ? You must have been ehausted before you get the answer. Therefore we need a more systematic way to do so. By our identity # and #, we recognize that the form ± y + y can be collapsed into ( ± y). (Here ± means + or.) Can it be done to our ? Sure. By comparing coefficients, we get = (Of course) and y = 7. (Take ± as ) But where does y, i.e. 49 come? The answer is easy: add nine and subtract nine. So: = + + = + ( ) = 7 9 This process is called Completing the square. And therefore: = 0 ( ) 7 9 = 0 ( ) 7 = 9 7 = 3 or 3 = 0 or 4 Sum and Product of Roots of Quadratic Equation We knew that a quadratic equation a + b + c could be factorized into ( α) ( β). We also knew that α and β are roots of the quadratic equation. Epand ( α) ( β), and by comparing coefficients, we get: b α + β = a c αβ = a This can help us find out sum and product of roots without solving the equation. After finding the sum and product, we can derive many other forms involving the two roots. Here gives some important forms composed of roots of a quadratic equation, α and β. For convenient, s denotes α + β, and p denotes αβ... α + β = + + α β αβ αβ s = α β p = α αβ + β = s 4 p (Assuming α > β) Quadratic Formula ( 二次方程公式 ) Completing the square is still quite inefficient. Actually, if we complete the square of a general quadratic equation, a + b + c = 0, we will get the equation solving a general quadratic equation, known as Quadratic formula : b b 4ac ± = a This equation is so important that should be memorized. 8

4 Discriminant of Quadratic Equation ( 二次方程之判別式 ) b ± b 4ac Recall the quadratic formula =. We see that the part inside the square root a sign, b 4ac is quite important to the nature of roots. If b 4ac > 0, we got two distinct real roots. If b 4ac = 0, we got one real root, or two repeated roots ( 重根 ). If b 4ac < 0, there is no real roots. Actually, b 4ac is called the Discriminant of a quadratic equation: to discriminate the nature of roots. It is often denoted as D or. Graph of Quadratic Equation To begin, see a graph of quadratic equation first: y y = ^ because = 0. After completing the square, the equation becomes y =, and 4 the minimum is? For a general quadratic equation, there is many useful information can be found on it. Let the equation be y = a + b + c, and the form after completing the square is y = a( h) + k. Then: a controls the shape of the graph. If a > 0, the graph opens upwards. If a < 0, the graph opens downwards. Moreover, the higher a is, the thinner the graph is. (0, c) is the y-intercept of the graph. (Discriminant) controls the relationship between the graph and the -ais. If > 0, the graph intersects the -ais at two points. If = 0, the graph touches, or tangent to ( 切於 ) the -ais at one point. If < 0, the graph does not intersect the -ais. (0, α), (0, β) are -intercepts of the graph. (α, β are roots) k is the etremum ( 極值 ) of the graph. Moreover, If a > 0, it is the minimum. If a < 0, it is the maimum. The point (h, k) is the verte ( 頂點 ) of the graph. Etension: The graph of quadratic equation is called a parabola ( 拋物線 ) Simultaneous Equations in Unknowns, One Linear, One Quadratic ( 聯立二元一次及二次方程 ) This is a graph of a quadratic equation, y = One can see it crosses the -ais at (-,0) and (-,0). Also, it crosses the y-ais at (0,). Anything else special? It achieves its minimum at ( 3, ). What is the fact behind these numbers? Let us uncover them. The 4 -intercepts are actually the roots, because y = 0 there. The y-intercept is the constant c, A set of simultaneous equations like: y = a + b + c y = m + k It can be solved by substituting () into (). Like quadratic equations, the set can have two solutions, one solution or no solutions. ( ) ( ) 9

5 Factor Theorem ( 因式定理 ) Remember the remainder theorem taught in Chapter? When a polynomial P() is divided by a, its remainder is P(a). But if P(a) = 0, P() is divisible by a, thus P() = ( a) Q() (Q() is the quotient), which implies a is a factor of P(). This gives the factor theorem: If P() is a polynomial and P(a) = 0, a is a factor of P(). For eample, to factorize P() = , we may employ the factor theorem and test for every possible a so that P(a) = 0. From our eperience in doing cross method for quadratic equation, we could also list out all factors of 6 and test for them one by one. a P() Thus, P() = ( + )( + )( + 3). Ecept testing out all possibilities, we may also test the possibilities until one zero is found (e.g. a = - above), and do synthetic division to get an equation with fewer degree (e.g here, as = ( + )( )) Once we got a quadratic equation left, instantly apply quadratic formula. A more complicated eample is that factorizing the polynomial P5 ( ) = Our list is quite long, 40 candidates. So, by trial and error, we got ( ) as one of the factors. Dividing, our new polynomial is 4 3 P4 ( ) = Still that 40 values to be tested, and we have ( + ) this time. ( ) 3 P3 = , and the factor is ( 4). What we left is only ( ) P = + 30, so apply quadratic formula to get ( 5) and ( + 6). Therefore, ( )( )( )( )( ) = Tedious, is it? When using factor theorem, you should first test the values that has a small absolute value. Remainder Theorem for Quadratic Divisor In Chapter Eercise Question 5, you proved that remainder theorem is true because P a = a a Q a + R = R. (Oh, I didn t know you haven t done that) ( ) ( ) ( ) By similar method, we can also deduce a remainder theorem whose divisor is quadratic. Assume the divisor is a + b + c, and dividend is P(). Thus, ( ) = ( + + ) ( ) + ( ) P a b c Q R Here, define R() = r + s. To make the Q() part disappear, one must have Q() = 0 or a + b + c = 0. We chose the latter one, as it is easier to implement. a + b + c = 0 when = α or = β, where α, β are roots. So we have got: P( α ) = R( α ) P( β ) = R( β ) Or P( α ) = rα + s P( β ) = rβ + s An SLE! P( α ) P( β ) Solving for r, we got r =. Substitute this into any of the equation above to get α β the value of s. By this, we got the remainder theorem for quadratic divisor. Revision: In this chapter, we ve learnt:. What is, how to, and why factorizing a polynomial. Solving quadratic equation 3. Sum and product of roots of quadratic equation 4. Discriminant of quadratic equation 5. Simultaneous equations in unknowns, one linear, one quadratic 6. Factor theorem 7. Remainder theorem for quadratic divisor 0

6 Eercise In the followings, if not specified, is the variable, k is a constant, and n is an integer.. Solve the following equations. (If there is no real roots, state that) a) = 0 b) = 0 c) = 0 d) + + = 0 e) = 0 f) = 0 + g) = + h) 3 =. Solve = 0 3. If α and β are roots of + + 3, find α 3 + β Factorize If + k + k = has repeated roots for, find k. 6. Refer to the followings. a) Find b) Find the remainder of (HKCEE 988) If 9 (k+) + = 0 has equal roots, a) Solve k. b) Solve the equation if k < (HKCEE 994) Solve ( 3) 3 = (HKCEE 996) Let α, β be real roots of λ + = 0, where λ >. Let s n = α n + β n, where n is a positive integer. a) Epress s and s 3 in terms of λ. b) Find a 5 λα 4 + α 3. c) Find s 5 λs 4 + s 3. d) It is known that s 3, s 4, s 5 0 and s 3 : s 4 : s 5 = 0 : 7λ : 5. Find λ. e) Find s 3. f) Evaluate y = 3 0. Solve 0 = y + 5. (ISMC 000 Final) If a b b = 0, find a b 00 [Hint: k 0 ]. (PCMSIMC 003, Modified) Let f() = a + b. If f(f()) = 4 + 9, a) Find all possible values of a and b. b) Find the smallest possible value of f(). 3. Assume A (k, k ), B (, k), C (k, ) are three points on the coordinates plane, where k. a) Epress the area of ABC in terms of k. b) Find the area of ABC when k = 3. c) Find the integral solution ( 整數解 ) of k when the area of ABC is (USAMO 995, Modified) If f() = a 4 b and f(-3) =, show that f() is not divisible by Prove the properties of sum and product of roots by quadratic formula. 6. (CaMO 99) Solve + = 3 + ( ) 7. Factorize [Hint: This is product of a quadratic and cubic equation How did cross method work?] 8. (ISMC 000 Final) Factorize z 6 + z 4 + z 3 + z +. [Hint: This is product of a quadratic and quartic equation]

7 Suggested Solutions for the Eercise a) 5, - b) 8, -7 c) ± 3 d) No real roots e),,, f) 0, 0, 5 ± 7 g) h) 7 a) a =, b = ; a =, b = b) 3a) 4 3 k k k + or ( k ) ( k + k + ) b) 6 c) 6) ± 5 7) ( + + 3) ( ) 8) (z z + ) (z 4 + z 3 + z + z + ) ), 5, 7 3) 0 4) ( ) ( ) ( 4) ( + 8) ( + ) 5) 6a) b) a) 5, -7 b) /3 8), 7 9a) s = λ ; s 3 = λ 3 3λ. b) 0 c) 0 d) 5 e) 5 f) ± ) = ; y = 4 8 ) 0