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\frac{16x^{2}-16x+3}{16}
Factor out \frac{1}{16}.
a+b=-16 ab=16\times 3=48
Consider 16x^{2}-16x+3. Factor the expression by grouping. First, the expression needs to be rewritten as 16x^{2}+ax+bx+3. To find a and b, set up a system to be solved.
-1,-48 -2,-24 -3,-16 -4,-12 -6,-8
Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 48.
-1-48=-49 -2-24=-26 -3-16=-19 -4-12=-16 -6-8=-14
Calculate the sum for each pair.
a=-12 b=-4
The solution is the pair that gives sum -16.
\left(16x^{2}-12x\right)+\left(-4x+3\right)
Rewrite 16x^{2}-16x+3 as \left(16x^{2}-12x\right)+\left(-4x+3\right).
4x\left(4x-3\right)-\left(4x-3\right)
Factor out 4x in the first and -1 in the second group.
\left(4x-3\right)\left(4x-1\right)
Factor out common term 4x-3 by using distributive property.
\frac{\left(4x-3\right)\left(4x-1\right)}{16}
Rewrite the complete factored expression.