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p+q=4 pq=4\left(-15\right)=-60
Factor the expression by grouping. First, the expression needs to be rewritten as 4b^{2}+pb+qb-15. To find p and q, set up a system to be solved.
-1,60 -2,30 -3,20 -4,15 -5,12 -6,10
Since pq is negative, p and q have the opposite signs. Since p+q is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -60.
-1+60=59 -2+30=28 -3+20=17 -4+15=11 -5+12=7 -6+10=4
Calculate the sum for each pair.
p=-6 q=10
The solution is the pair that gives sum 4.
\left(4b^{2}-6b\right)+\left(10b-15\right)
Rewrite 4b^{2}+4b-15 as \left(4b^{2}-6b\right)+\left(10b-15\right).
2b\left(2b-3\right)+5\left(2b-3\right)
Factor out 2b in the first and 5 in the second group.
\left(2b-3\right)\left(2b+5\right)
Factor out common term 2b-3 by using distributive property.
4b^{2}+4b-15=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
b=\frac{-4±\sqrt{4^{2}-4\times 4\left(-15\right)}}{2\times 4}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
b=\frac{-4±\sqrt{16-4\times 4\left(-15\right)}}{2\times 4}
Square 4.
b=\frac{-4±\sqrt{16-16\left(-15\right)}}{2\times 4}
Multiply -4 times 4.
b=\frac{-4±\sqrt{16+240}}{2\times 4}
Multiply -16 times -15.
b=\frac{-4±\sqrt{256}}{2\times 4}
Add 16 to 240.
b=\frac{-4±16}{2\times 4}
Take the square root of 256.
b=\frac{-4±16}{8}
Multiply 2 times 4.
b=\frac{12}{8}
Now solve the equation b=\frac{-4±16}{8} when ± is plus. Add -4 to 16.
b=\frac{3}{2}
Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.
b=-\frac{20}{8}
Now solve the equation b=\frac{-4±16}{8} when ± is minus. Subtract 16 from -4.
b=-\frac{5}{2}
Reduce the fraction \frac{-20}{8} to lowest terms by extracting and canceling out 4.
4b^{2}+4b-15=4\left(b-\frac{3}{2}\right)\left(b-\left(-\frac{5}{2}\right)\right)
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and -\frac{5}{2} for x_{2}.
4b^{2}+4b-15=4\left(b-\frac{3}{2}\right)\left(b+\frac{5}{2}\right)
Simplify all the expressions of the form p-\left(-q\right) to p+q.
4b^{2}+4b-15=4\times \frac{2b-3}{2}\left(b+\frac{5}{2}\right)
Subtract \frac{3}{2} from b by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.
4b^{2}+4b-15=4\times \frac{2b-3}{2}\times \frac{2b+5}{2}
Add \frac{5}{2} to b by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
4b^{2}+4b-15=4\times \frac{\left(2b-3\right)\left(2b+5\right)}{2\times 2}
Multiply \frac{2b-3}{2} times \frac{2b+5}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
4b^{2}+4b-15=4\times \frac{\left(2b-3\right)\left(2b+5\right)}{4}
Multiply 2 times 2.
4b^{2}+4b-15=\left(2b-3\right)\left(2b+5\right)
Cancel out 4, the greatest common factor in 4 and 4.
x ^ 2 +1x -\frac{15}{4} = 0
Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 4
r + s = -1 rs = -\frac{15}{4}
Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C
r = -\frac{1}{2} - u s = -\frac{1}{2} + u
Two numbers r and s sum up to -1 exactly when the average of the two numbers is \frac{1}{2}*-1 = -\frac{1}{2}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>
(-\frac{1}{2} - u) (-\frac{1}{2} + u) = -\frac{15}{4}
To solve for unknown quantity u, substitute these in the product equation rs = -\frac{15}{4}
\frac{1}{4} - u^2 = -\frac{15}{4}
Simplify by expanding (a -b) (a + b) = a^2 – b^2
-u^2 = -\frac{15}{4}-\frac{1}{4} = -4
Simplify the expression by subtracting \frac{1}{4} on both sides
u^2 = 4 u = \pm\sqrt{4} = \pm 2
Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u
r =-\frac{1}{2} - 2 = -2.500 s = -\frac{1}{2} + 2 = 1.500
The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.