Question 45127
{{{2x^2+7x+3=(ax+b)(cx+d)}}}
looking at the coefficient of {{{x^2}}} on the left hand side of the equation:
{{{2x^2=(ax)(bx)}}}
{{{2x^2=abx^2}}}
So {{{ac=2}}}........................(1)
Now look at the cofactors of x:
{{{7x=adx+bcx}}}
So {{{7=ad+bc}}}......................(2)
Now look at the constant terms:
{{{3=bd}}}..............................(3)
Looking at equation (1), since thwe number 2 is a prime number, there are only two numbers that "a" and "c" could be: 1 and 2.
Let a=2 and c=1.
Looking at equation (3), since thwe number 3 is a prime number, there are only two numbers that "b" and "d" could be: 1 and 3. This time we need to make sure that we get these two numbers the right way round, so we check with equation (2):
{{{7=2d+b}}}
so if b and d can omly be 1 or 2, the only the above equation works is if b=1 and d=3.
Put the values of a, b, c and d into the equation at the very top:
{{{2x^2+7x+3=(2x+1)(x+3)}}}

I hope this helps.
Adam
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