An expression in the form Ax2+Bx+C is called a quadratic expression. It can be factorized as (ax+b)(cx+d). Generally in examinations all the constants A,B,C,a,b,c,d are kept integers. The simple process to to factorize the expression is to write the term Bx as sum of two terms ie. Bx=b'x+b"x. It makes the expression in four terms and then take commons out from first two terms and last two terms.
Ax2+Bx+C= Ax2+b'x+b"x+C= (Ax2+b'x)+(b"x+C)
What remains inside the two pairs of brackets is same linear expressions (Only if b' and b" is chosen judiciously) which are again taken out as commons, thus factorizing the expression easily in simple way. Let us see an example.
6x2+11x+3 = 6x2+9x+2x+3 = 3x(2x+3) +(2x+3) = (2x+3)(3x+1)
But how did we know that breaking 11x in 9x and 2x will give us a common factor as 2x+3. Why did not we break in pairs as 6x,5x; 7x,4x; 8x,3x or 10x,x.
What did we do that first we remembered the product of A and C. In this example 6*3=18. Then we picked that pair of b' and b" which product was same. Here 9 and 2 give 18 on multiplication, so we picked this pair. But we must remember that algebraic sum of b'x and b"x should be equal to Bx. We face different situations depending upon the signs preceding B and C. We always keep A positive, even if A is -A the negative is kept outside keeping whole expression inside a bracket.
Case I: When all terms are positive
The process is same as in example above. Here numerical value of b'<B and b"<B
Case II: When B is positive but C is negative
As in expression Ax2+Bx-C. Since multiplication of first and last term is negative, we will have to break B in such a way that the product b'*b" is also negative. It can only be achieved if one of them is positive and the other negative. Let it be b' and -b" ie B=b'-b". Since B is positive, numerical value of b'>b". See in example below.
2x2+x-3 Product of 2 and -3 is -6. So we have to break x in a pair such that their difference is x but product is -6x2 , So start the pair as 2x,-x; 3x,-2x ...... Here we get the desired pair in 2nd trial only ie 3x,-2x which product is -6 x2 . So we proceed as thus,
2x2+x-3= 2x2+3x-2x-3=x(2x+3)-(2x+3)=(2x+3)(x-1)
Case III: When B and C both are negative
As in expression Ax2-Bx-C. The process will be similar to Case-II except that numerical value of b'<b" which will give B negative in B=b'-b". See the example below.
3x2-2x-8 Product of 3 and -8 is -24. So we break mid term such that difference of their numerical value is 2 and product is 24. We proceed as 1,-3; 2,-4; 3,-5; 4,-6 ... The 4th trial gives us the desired result. Now the factorization is simple as we write,
3x2+4x-6x-8= x(3x+4)-2(3x+4) =(3x+4)(x-2)
Case IV: When B is negative and C is Positive
As in expression Ax2-Bx+C. The process is similar to Case-I. Here too numerical value of b'<B and b"<B. Since sign before both b' and b" is negative their product is positive as is the sign of product of first and last term. Let us see the example below.
4x2-15x+9
Product of 4 and 9 is 36. So we break 15 in a pair such that sum of numerical value of the pair is 15 and product is 36. We start as 14,1; 13,2; 12,3 ...... So in the 3rd trial we get the desired result. Our pair is 12,3. We keep sign before them negative so that their product is positive. Now it is simple to factorize as below.
4x2-12x-3x+9 = 4x(x-3)-3(x-3) =(x-3)(4x-3)
By practice guessing the pairs in these four cases gets easy. It is very helpful in quickly solving the quadratic equations for their roots.
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