Factorising Quadratics with coefficient a not equal to 1
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)
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