Algebra InputsTrigonometry InputsCalculus InputsMatrix Inputs Basic algebra trigonometry calculus statistics matrices CharactersFactor \left(3c-4\right)\left(c+3\right)Solution StepsSteps Using the Quadratic FormulaSteps Using Direct Factoring MethodView solution stepsSolution Steps 3 c ^ { 2 } + 5 c - 12 Factor the expression by grouping. First, the expression needs to be rewritten as 3c^{2}+ac+bc-12. To find a and b, set up a system to be solved. a+b=5 ab=3\left(-12\right)=-36 Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -36. -1,36 -2,18 -3,12 -4,9 -6,6 Calculate the sum for each pair. -1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0 The solution is the pair that gives sum 5. a=-4 b=9 Rewrite 3c^{2}+5c-12 as \left(3c^{2}-4c\right)+\left(9c-12\right). \left(3c^{2}-4c\right)+\left(9c-12\right) Factor out c in the first and 3 in the second group. c\left(3c-4\right)+3\left(3c-4\right) Factor out common term 3c-4 by using distributive property. \left(3c-4\right)\left(c+3\right) Evaluate \left(3c-4\right)\left(c+3\right)QuizPolynomial5 problems similar to: 3 c ^ { 2 } + 5 c - 12
Similar Problems from Web Search
3c^2-5c-12http://www.tiger-algebra.com/drill/3c~2-5c-12/ 3c2-5c-12 Final result : (c - 3) • (3c + 4) Step by step solution : Step 1 :Equation at the end of step 1 : (3c2 - 5c) - 12 Step 2 :Trying to factor by splitting the middle term ... c^2+5c-14http://www.tiger-algebra.com/drill/c~2_5c-14/ c2+5c-14 Final result : (c + 7) • (c - 2) Step by step solution : Step 1 :Trying to factor by splitting the middle term 1.1 Factoring c2+5c-14 The first term is, c2 its coefficient is ... 3c^2+2c-16https://www.tiger-algebra.com/drill/3c~2_2c-16/ 3c2+2c-16 Final result : (c - 2) • (3c + 8) Step by step solution : Step 1 :Equation at the end of step 1 : (3c2 + 2c) - 16 Step 2 :Trying to factor by splitting the middle term ... 12c^2+5c-2https://www.tiger-algebra.com/drill/12c~2_5c-2/ 12c2+5c-2 Final result : (4c - 1) • (3c + 2) Step by step solution : Step 1 :Equation at the end of step 1 : ((22•3c2) + 5c) - 2 Step 2 :Trying to factor by splitting the middle term ... c^2+5c-66=0http://www.tiger-algebra.com/drill/c~2_5c-66=0/ c2+5c-66=0 Two solutions were found : c = 6 c = -11 Step by step solution : Step 1 :Trying to factor by splitting the middle term 1.1 Factoring c2+5c-66 The first term is, c2 its ... 12k^2-36k+27=0https://www.tiger-algebra.com/drill/12k~2-36k_27=0/ 12k2-36k+27=0 One solution was found : k = 3/2 = 1.500 Step by step solution : Step 1 :Equation at the end of step 1 : ((22•3k2) - 36k) + 27 = 0 Step 2 : Step 3 :Pulling ...More Items
Share
CopyCopied to clipboarda+b=5 ab=3\left(-12\right)=-36 Factor the expression by grouping. First, the expression needs to be rewritten as 3c^{2}+ac+bc-12. To find a and b, set up a system to be solved.-1,36 -2,18 -3,12 -4,9 -6,6 Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -36.-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0 Calculate the sum for each pair.a=-4 b=9 The solution is the pair that gives sum 5.\left(3c^{2}-4c\right)+\left(9c-12\right) Rewrite 3c^{2}+5c-12 as \left(3c^{2}-4c\right)+\left(9c-12\right).c\left(3c-4\right)+3\left(3c-4\right) Factor out c in the first and 3 in the second group.\left(3c-4\right)\left(c+3\right) Factor out common term 3c-4 by using distributive property.3c^{2}+5c-12=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.c=\frac{-5±\sqrt{5^{2}-4\times 3\left(-12\right)}}{2\times 3} 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.c=\frac{-5±\sqrt{25-4\times 3\left(-12\right)}}{2\times 3} Square 5.c=\frac{-5±\sqrt{25-12\left(-12\right)}}{2\times 3} Multiply -4 times 3.c=\frac{-5±\sqrt{25+144}}{2\times 3} Multiply -12 times -12.c=\frac{-5±\sqrt{169}}{2\times 3} Add 25 to 144.c=\frac{-5±13}{2\times 3} Take the square root of 169.c=\frac{-5±13}{6} Multiply 2 times 3.c=\frac{8}{6} Now solve the equation c=\frac{-5±13}{6} when ± is plus. Add -5 to 13.c=\frac{4}{3} Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.c=-\frac{18}{6} Now solve the equation c=\frac{-5±13}{6} when ± is minus. Subtract 13 from -5.c=-3 Divide -18 by 6.3c^{2}+5c-12=3\left(c-\frac{4}{3}\right)\left(c-\left(-3\right)\right) Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{4}{3} for x_{1} and -3 for x_{2}.3c^{2}+5c-12=3\left(c-\frac{4}{3}\right)\left(c+3\right) Simplify all the expressions of the form p-\left(-q\right) to p+q.3c^{2}+5c-12=3\times \frac{3c-4}{3}\left(c+3\right) Subtract \frac{4}{3} from c by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.3c^{2}+5c-12=\left(3c-4\right)\left(c+3\right) Cancel out 3, the greatest common factor in 3 and 3.x ^ 2 +\frac{5}{3}x -4 = 0Quadratic 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 3r + s = -\frac{5}{3} rs = -4Let 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 = Cr = -\frac{5}{6} - u s = -\frac{5}{6} + uTwo numbers r and s sum up to -\frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*-\frac{5}{3} = -\frac{5}{6}. 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{5}{6} - u) (-\frac{5}{6} + u) = -4To solve for unknown quantity u, substitute these in the product equation rs = -4\frac{25}{36} - u^2 = -4Simplify by expanding (a -b) (a + b) = a^2 – b^2-u^2 = -4-\frac{25}{36} = -\frac{169}{36}Simplify the expression by subtracting \frac{25}{36} on both sidesu^2 = \frac{169}{36} u = \pm\sqrt{\frac{169}{36}} = \pm \frac{13}{6} Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable ur =-\frac{5}{6} - \frac{13}{6} = -3 s = -\frac{5}{6} + \frac{13}{6} = 1.333The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
Examples
Quadratic equation { x } ^ { 2 } - 4 x - 5 = 0Trigonometry 4 \sin \theta \cos \theta = 2 \sin \thetaLinear equation y = 3x + 4Arithmetic 699 * 533Matrix \left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]Simultaneous equation \left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.Differentiation \frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }Integration \int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d xLimits \lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}Back to top
Algebra Inputs
Trigonometry Inputs
Calculus Inputs
Matrix InputsSolvePracticePlay
Solution StepsSteps Using the Quadratic FormulaSteps Using Direct Factoring MethodView solution stepsSolution Steps 3 c ^ { 2 } + 5 c - 12 Factor the expression by grouping. First, the expression needs to be rewritten as 3c^{2}+ac+bc-12. To find a and b, set up a system to be solved. a+b=5 ab=3\left(-12\right)=-36 Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -36. -1,36 -2,18 -3,12 -4,9 -6,6 Calculate the sum for each pair. -1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0 The solution is the pair that gives sum 5. a=-4 b=9 Rewrite 3c^{2}+5c-12 as \left(3c^{2}-4c\right)+\left(9c-12\right). \left(3c^{2}-4c\right)+\left(9c-12\right) Factor out c in the first and 3 in the second group. c\left(3c-4\right)+3\left(3c-4\right) Factor out common term 3c-4 by using distributive property. \left(3c-4\right)\left(c+3\right) Evaluate \left(3c-4\right)\left(c+3\right)
Back to top 
