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Learn the different methods of factoring quadratic polynomials, including grouping terms, using identities, and completing the square. Practice examples provided.
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CHAPTER 3- MORE ABOUT FACTORIZATION expand x2 x2 +xy -xy +xy - xy +y2 +y2 -xy +xy x2 -y2 factorize Basic knowledge Factors of 30 = 1,2,3,5,6,10,15 and 30 3 important identities (x+y)2(x+y)(x+y) =x2+2xy+y2 =x2-2xy+y2 (x-y)2(x -y)( x -y) =x2-y2 (x+y)(x-y)
Expansion product of 2 linear polynomials (2x+3)(x-1)=2x2-2x+3x-3 =2x2+x-3 3.1 factorization of polynomials Changing a polynomial into the product of its factors. Basic steps: Taking out each term of a polynomial Example 1 Factorize the following expressions (a) 5ab+10ac=5a(b+2c) (b) 8x2-4x=4x(2x-1) (c) 8x-8+2x-5 =2x-8(4+x3)
Classwork 3.1 Factorize the following expressions (1) 3xy+6yz=3y(x+2z) (2) 9x2-9x=9x(x-1) (3) 4x2y-6x2 =2x2(2y-3) (4) 9x2yz-3xy2z=3xyz(3x-y) Grouping terms for factorization Example 2 Factorize the following expressions (a) ax+ay+bx+by=a(x+y)+b(x+y)=(a+b)(x+y) (b) 5p2+3p-5pq-3q=p(5p+3)-q(5p+3)=(p-q)(5p+3) (c) 3b2-2cd-2bc+3bd=3b2-2bc +3bd -2cd=b(3b-2c)+d(3b-2c) =(b+d)(3b-2c)
Classwork 3.2 Factorize the following expressions (a) 3a+3b+ax+bx =3(a+b)+x(a+b)= (3+x)(a+b) (b) a2(2-x)-7(x-2) = a2(2-x)+7(2-x) =(a2+7)(2-x) (c) x2+yz+xz+xy =x2+xz+xy+yz=x(x+z)+y(x+z)=(x+y)(x+z) (d) xy-xz+wz-wy =x(y-z)+w(z-y)=x(y-z)-w(y-z)=(x-w)(y-z) = 5c3 +15cde+ac2 +3ade (e) 5c3+3ade+15cde+ac2 = 5c(c2 +3de)+a (c2+3de) = (5c+a)(c2 +3de) (f) 4px-qy+2qx-2py= 4px +2qx -2py -qy=2x(2p+q)-y(2p+q) =(2x-y)(2p+q)
Using identities for factorization Factorize the following expressions Rule 1 a2+2ab+b2 (a+b)2 Rule 2 a2-2ab+b2 (a-b)2 (a+b)(a-b) a2-b2 Rule 3 Factorize the following expressions (a) 25m2-16n2=(5m)2-(4n)2 =(5m+4n)(5m-4n) Rule 3 Rule 3 (b) x4-81=(x2)2-92 =(x2+9)(x2-9)= (x2+9)(x+3)(x-3) Take out common term (c) 5a(b+c)2-180a3=5a[(b+c)2-36a3] Rule 3 =5a[(b+c)2-(6a)3] =5a (b+c+6a) (b+c-6a)
Factorize the following expressions (1) 49x2-4y2=(7x)2-(2y)2 =(7x+2y)(7x-2y) Rule 3 Rule 3 (2) 4x4-64=4(x4-16)=4(x+4)(x-4) (3) x4-y4 =(x2)2-(y2)2=(x2-y2)(x2+y2)=(x+y)(x-y)(x2+y2) (4) 7x(x+y)2-112xz2=7x[(x+y)2-16z2]=7x[(x+y)2-(4z)2]=7x(x+y+4z)(x+y+4z)
Example 4 -Factorize the following expressions (a+b)2 (a-b)2 (a-b)2 (a+b)2 (a-b)2 a2+2ab+b2 a2-2ab+b2 a2-2ab+b2 a2+2ab+b2 a2-2ab+b2 Rule 1 Rule2 Rule2 Rule 1 Rule2 (a+b)(a-b) a2-b2 Rule 3 (a) 4m2+4m+1 = (2m)2 +2(2m)(1)+12 =(2m+1)2 (b)2z2-12z+18 = 2(z2-6z+9)=2[z2–2(z)(3)+32]=2(z-3)2 (c)49y2-56yz+16z2=(7y)2-2(7y)(4z)+(4z)2 =(7y-4z)2 (d)a2+12ab+36b2-25c2+10c-1 =(a2+12ab+36b2)-(25c2-10c+1) =[a2+2(a)(6b)+(6b)2]-[(5c)2-2(5c)(1)+12] =(a+6b)2-(5c-1)2 =(a+6b+5c-1)(a+6b-5c+1)
Classwork 3.4 -Factorize the following expressions (a+b)2 (a-b)2 (a-b)2 (a+b)2 (a+b)2 a2-2ab+b2 a2-2ab+b2 a2+2ab+b2 a2+2ab+b2 a2+2ab+b2 Rule2 Rule 1 Rule 1 Rule 1 Rule2 (a+b)(a-b) a2-b2 Rule 3 (1) x2-4xy+4y2 = x2 -2(x)(2y)+(2y)2 =(x-2y)2 (2)6a2+12a+6 = 6(a2+2a+1)=6[a2+2(a)(1)+12]=6(a+1)2 (3)(x-y)2+2(x-y)+1= (x-y)2+2(x-y)(1)+12=(x-y+1)2 (4)9x2-12xy+4y2-1-4z-4z2 =(9x2-12xy+4y2)-(12+4z+4z2) =[(3x)2-2(3x)(2y)+(2y)2]-[(1)2+2(2z)(1)+(2z)2] =(3x-2y)2-(1+2z)2 =(3x-2y+1+2z)(3x-2y-1-2z)
3.2 FACTORIZATION OF QUADRATIC POLYYNOMIALS IN ONE VARIABLE + 6 + 1 + 3 + 2 Quadratic polynomial – one variable, highest power =2 e.g.1x2+5x +63x2 - 2x+ 3 Ax2+Bx+C where A,B,C are constant A not zero. If it is ,the polynomial is Bx+C, not quadratic =(x ) (x ) 1x2+5x +6
(x+p)(x+q) = x2 +px+qz+pq =1x2 +(p+q)x+pq - - 8 1 + + 2 4 =ax2 + b x + c If we can find 2 numbers p and q , such that p + q= b, pq= c Then quadratic polynomial can be factorized. =( x ) ( x ) x2+6x +8
- - 1 15 + + 3 5 =( x ) ( x ) x2+8x +15
- - 1 18 + + 2 9 =( x ) ( x ) x2+11x +18
(x+p)(x+q) = x2 +px+qz+pq = x2 +(p+q)x+pq =ax2 + b x + c If we can find 2 numbers p and q , such that p + q= b pq = c Then quadratic polynomial can be factorized.
- - 24 1 + - + + 8 12 3 2 Example 5 (a) =( x ) ( x ) x2-5x -24
- - 6 1 + - + + -5x+x2 -6 (b) =( x ) ( x ) x2 -5x -6
15x-54-x2 - - 54 1 + - + + 18 9 27 2 6 3 -x2+ 15x -54 (c) -(x2-15x +54)= -( x ) ( x )
3.3FACTORIZATION OF QUADRATIC POLYYNOMIALS IN ONE VARIABLE + - 4 4 4 4 - + 8 16 8 16 1 1 2 2 (a) 3x2-2x -16 =( 3x ) ( x )
3.3FACTORIZATION OF QUADRATIC POLYYNOMIALS IN ONE VARIABLE + - 4 4 - + 16 8 8 16 1 1 2 2 9x2-6x –48= 3(3x2-2x –16) (b) =3 ( 3x ) ( x )
- + 12 1 + - 6 3 4 3 1 1 2 4 2 3 3 12 12 6 2 6 4 4 (c) 11x -5x2+12= -5x2 + 11x +12 = -(5x2 -11x -12) = - ( 5x ) ( x )
FORM 4 Ax2+Bx +C 3x2-2x –16=0 or or
COMPLETING SQUARE X2+2YX+Y2=(X+Y)2 X2+6X+? = X2+2(3)X+32= X2+6X+9= (X+3)2
2X2-8X-24= 2(X2-4X-12)= 2(X+2)(X-6) 4 4 = - + - - 2 2 2 2 ( X 4 X ( ) ( ) 12) 2 2 = - + - - 2 2 ( X 4 X 4 4 12 ) = - - 2 2 [( X 2 ) 16 ] = - - 2 2 2 [( X 2 ) 4 ] Another way to factorize complicated polynomials 2X2-8X-24= 2(X2-4X-12)