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In fact 2 and 10 do that 2 x 10 = 20 and 2 + 10 = 12 Factorise: 2x2+12x+10 Step 1: Find two numbers that multiply to give a x c, and add to give b In this example a x c = 2 x 10 = 20 and b = 12,so we want two numbers that multiply to give 20and add up to give 12. 2x2+12x+10=2x2+2x+10x+10=2x(x+1)+10(x+1) Step 2: Use these numbers to rewrite the middle term. Factorise: 2x2+12x+10 Step 3: Factorise the first two terms 2x2+2x and the last two terms 10x+10 seperately so 2x2+2x=2x(x+1) and 10x+10=10(x+1) 2x2+12x+10 = 2x2+2x+10x+10 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (x+1). We can take the common factor in front of the bracket. From Step 3 we know that: 2x2+12x+10=2x2+2x+10x+10=2x(x+1)+10(x+1) Factorise: 2x2+12x+10 so we finally factorised the expression in question: 2x(x+1)+10(x+1) = (x+1)(2x+10) 2x2+12x+10 = (x+1)(2x+10) In fact 4 and 6 do that 4 x 6 = 24 and 4 + 6 = 10 Factorise: 2x2+10x+12 Step 1: Find two numbers that multiply to give a x c, and add to give b In this example a x c = 2 x 12 = 24 and b = 10,so we want two numbers that multiply to give 24and add up to give 10. 2x2+10x+12=2x2+4x+6x+12=2x(x+2)+6(x+2) Step 2: Use these numbers to rewrite the middle term. Factorise: 2x2+10x+12 Step 3: Factorise the first two terms 2x2+4x and the last two terms 6x+12 seperately so 2x2+4x=2x(x+2) and 6x+12=6(x+2) 2x2+10x+12 = 2x2+4x+6x+12 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (x+2). We can take the common factor in front of the bracket. From Step 3 we know that: 2x2+10x+12=2x2+4x+6x+12=2x(x+2)+6(x+2) Factorise: 2x2+10x+12 so we finally factorised the expression in question: 2x(x+2)+6(x+2) = (x+2)(2x+6) 2x2+10x+12 = (x+2)(2x+6) In fact 3 and 12 do that 3 x 12 = 36 and 3 + 12 = 15 Factorise: 3x2+15x+12 Step 1: Find two numbers that multiply to give a x c, and add to give b In this example a x c = 3 x 12 = 36 and b = 15,so we want two numbers that multiply to give 36and add up to give 15. 3x2+15x+12=3x2+3x+12x+12=3x(x+1)+12(x+1) Step 2: Use these numbers to rewrite the middle term. Factorise: 3x2+15x+12 Step 3: Factorise the first two terms 3x2+3x and the last two terms 12x+12 seperately so 3x2+3x=3x(x+1) and 12x+12=12(x+1) 3x2+15x+12 = 3x2+3x+12x+12 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (x+1). We can take the common factor in front of the bracket. From Step 3 we know that: 3x2+15x+12=3x2+3x+12x+12=3x(x+1)+12(x+1) Factorise: 3x2+15x+12 so we finally factorised the expression in question: 3x(x+1)+12(x+1) = (x+1)(3x+12) 3x2+15x+12 = (x+1)(3x+12) Step 4: The common factor for these two terms above is (x+4). Take it in front of the bracket. You could also split the 15x into 12x+3x instead. 3x2+15x+12=3x2+12x+3x+12=3x(x+4)+3(x+4) Factorise: 3x2+15x+12 so we finally factorised the expression in question: 3x(x+4)+3(x+4) = (x+4)(3x+3) 3x2+15x+12 = (x+4)(3x+3) In fact 2 and 12 do that 2 x 12 = 24 and 2 + 12 = 15 Factorise: 4x2+14x+6 Step 1: Find two numbers that multiply to give a x c, and add to give b In this example a x c = 4 x 6 = 24 and b = 14,so we want two numbers that multiply to give 24and add up to give 14. 4x2+14x+6=4x2+2x+12x+6=2x(2x+1)+6(2x+1) Step 2: Use these numbers to rewrite the middle term. Factorise: 4x2+14x+6 Step 3: Factorise the first two terms 4x2+2x and the last two terms 12x+6 seperately so 4x2+2x=2x(2x+1) and 12x+6=6(2x+1) 4x2+14x+6 = 4x2+2x+12x+6 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (2x+1). Let's take the common factor in front of the bracket. From Step 3 we know that: 4x2+14x+6=4x2+2x+12x+6=2x(2x+1)+6(2x+1) Factorise: 4x2+14x+6 so we finally factorised the expression in question: 2x(2x+1)+6(2x+1) = (2x+1)(2x+6) 4x2+14x+6 = (2x+1)(2x+6) So 15 and -8 do that 15x(-8) =-120 and 15+(-8) = 7 Swap order and add minus sign Factorise: 6x2+7x-20 In this example a x c = 6 x (-20) = -120 and b = 7,so we want two numbers that multiply to give -120and add up to give 7. One of the number needs to bepositive and the other number needs to be negative. Factors of 120: Step 1: Find two numbers that multiply to give a x c, and add to give b -120,-60,-40,-30,-24,-20,-15,-12,-10, -8, -6, -5, -4, -3, -2, -1 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120 Step 2: Use these numbers to rewrite the middle term. Factorise: 6x2+7x-20 Step 3: Factorise the first two terms 6x2+15x and the last two terms -8x-20 seperately so 6x2+7x-20=6x2+15x-8x-20=3x(2x+5)-4(2x+5) 6x2+15x=3x(2x+5) and -8x-20=-4(2x+5) 6x2+7x-20 = 6x2+15x-8x-20 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (2x+5). Let's take the common factor in front of the bracket. From Step 3 we know that: 6x2+7x-20=6x2+15x-8x-20=3x(2x+5)-4(2x+5) Factorise: 6x2+7x-20 so we finally factorised the expression in question: 3x(2x+5)-4(2x+5) = (2x+5)(3x-4) 6x2+7x-20 = (2x+5)(3x-4) Factorise: 4x2+10x+6 Step 1: Find two numbers that multiply to give a x c, and add to give b In fact 4 and 6 do that 4 x 6 = 24 and 4 + 6 = 10 In this example a x c = 4 x 6 = 24 and b = 10,so we want two numbers that multiply to give 24and add up to give 10. Step 2: Use these numbers to rewrite the middle term. Factorise: 4x2+10x+6 Step 3: Factorise the first two terms 4x2+4x and the last two terms 6x+6 seperately so 4x2+10x+6=4x2+4x+6x+6=4x(x+1)+6(x+1) 4x2+4x=4x(x+1) and 6x+6=6(x+1) 4x2+10x+6 = 4x2+4x+6x+6 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (x+1). Let's take the common factor in front of the bracket. From Step 3 we know that: 4x2+10x+6=4x2+4x+6x+6=4x(x+1)+6(x+1) Factorise: 4x2+10x+6 so we finally factorised the expression in question: 4x(x+1)+6(x+1) = (x+1)(4x+6) 4x2+10x+6 = (x+1)(4x+6) In fact 5 and 12 do that 5 x 12 = 60 and 5 + 12 = 17 Factorise: 3x2+17x+20 Step 1: Find two numbers that multiply to give a x c, and add to give b In this example a x c = 3 x 20 = 60 and b = 17,so we want two numbers that multiply to give 60and add up to give 17. 3x2+17x+20=3x2+5x+12x+20=x(3x+5)+4(3x+5) Step 2: Use these numbers to rewrite the middle term. Factorise: 3x2+17x+20 Step 3: Factorise the first two terms 3x2+5x and the last two terms 12x+20 seperately so 3x2+5x=x(3x+5) and 12x+20=4(3x+5) 3x2+17x+20 = 3x2+5x+12x+20 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (3x+5). Let's take the common factor in front of the bracket. From Step 3 we know that: 3x2+17x+20=3x2+5x+12x+20=x(3x+5)+4(3x+5) Factorise: 3x2+17x+20 so we finally factorised the expression in question: x(3x+5)+4(3x+5) = (3x+5)(x+4) 3x2+17x+20 = (3x+5)(x+4) In fact -4 and -6 do that -4x-6=24 and -4+(-6)=-10 Factorise: 3x2-10x+8 Step 1: Find two numbers that multiply to give a x c, and add to give b In this example a x c = 3 x 8 = 24 and b = -10,so we want two numbers that multiply to give 24and add up to give -10. Both of these numbers needto be negative numbers then. Step 2: Use these numbers to rewrite the middle term. Factorise: 3x2-10x+8 Step 3: Factorise the first two terms 3x2-6x and the last two terms -4x+8 seperately so 3x2-10x+8=3x2-6x-4x+8=3x(x-2)-4(x-2) 3x2-6x=3x(x-2) and -4x+8=-4(x-2) 3x2-10x+8 = 3x2-6x-4x+8 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (x-2). Let's take the common factor in front of the bracket. From Step 3 we know that: Factorise: 3x2-10x+8 so we finally factorised the expression in question: 3x2-10x+8=3x2-6x-4x+8=3x(x-2)-4(x-2) 3x(x-2)-4(x-2) = (x-2)(3x-4) 3x2-10x+8 = (x-2)(3x-4) Factorise: 4x2-10x-14 In this example a x c = 4 x -14 = -56 and b = -10,so we want two numbers that multiply to give -56and add up to give -10. One of the number needs tobe negative and the other needs to be positive. Step 1: Find two numbers that multiply to give a x c, and add to give b 4 and -14 do that 4x-14=-56 and 4+(-14)=-10 Step 2: Use these numbers to rewrite the middle term. Factorise: 4x2-10x-14 Step 3: Factorise the first two terms 4x2+4x and the last two terms -14x-14 seperately so 4x2-10x-14=4x2+4x-14x-14=4x(x+1)-14(x+1) 4x2+4x=4x(x+1) and -14x-14=-14(x+1) 4x2-10x-14 = 4x2+4x-14x-14 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (x+1). Let's take the common factor in front of the bracket. From Step 3 we know that: Factorise: 4x2-10x-14 so we finally factorised the expression in question: 4x2-10x-14=4x2+4x-14x-14=4x(x+1)-14(x+1) 4x(x+1)-14(x+1) = (x+1)(4x-14) 4x2-10x-14 = (x+1)(4x-14) Factorise: 4x2+7x-15 12 and -5 do that 12 x -5 = -60 and 12 + (-5) = 7 Step 1: Find two numbers that multiply to give a x c, and add to give b In this example a x c = 4 x -15 = -60 and b = 7,so we want two numbers that multiply to give -60and add up to give 7. One of the number needs tobe negative and the other needs to be positive. Step 2: Use these numbers to rewrite the middle term. Factorise: 4x2+7x-15 Step 3: Factorise the first two terms 4x2+12x and the last two terms -5x-15 seperately so 4x2+7x-15=4x2+12x-5x-15=4x(x+3)-5(x+3) 4x2+12x=4x(x+3) and -5x-15=-5(x+3) 4x2+7x-15 = 4x2+12x-5x-15 Step 4: Now we get clearly visible common factor for these two terms above. The common factor here is (x+3). Let's take the common factor in front of the bracket. From Step 3 we know that: Factorise: 4x2+7x-15 so we finally factorised the expression in question: 4x2+7x-15=4x2+12x-5x-15=4x(x+3)-5(x+3) 4x(x+3)-5(x+3) = (x+3)(4x-5) 4x2+7x-15 = (x+3)(4x-5) |