2x^2-5x-3 how to factorise this
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2x^2--5x--3
=2x^2--6x+1x--3
=2x(x--3)+1(x--3)
=(2x+1)(x--3)
=2x^2--6x+1x--3
=2x(x--3)+1(x--3)
=(2x+1)(x--3)
Anonymous:
hello
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the quadratic equation is of the form "ax²-bx+c"
now for this to factorise, multiply a and c, and then try this multiplication ac to break other in the form of submission or addition which will be equals to b.
now in this case a = 2, b=5, c=-3
multiplying 'a' and 'c' we get "-6" as "ac", and now to break this 'ac' = b; that is "-5 = -6 + 1".
so 2 x² - 5 x - 3 = 0
can be written as
→ 2x² - 6x + x - 3 = 0
and then solved...
now for this to factorise, multiply a and c, and then try this multiplication ac to break other in the form of submission or addition which will be equals to b.
now in this case a = 2, b=5, c=-3
multiplying 'a' and 'c' we get "-6" as "ac", and now to break this 'ac' = b; that is "-5 = -6 + 1".
so 2 x² - 5 x - 3 = 0
can be written as
→ 2x² - 6x + x - 3 = 0
and then solved...
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