Algebra InputsTrigonometry InputsCalculus InputsMatrix Inputs Basic algebra trigonometry calculus statistics matrices CharactersFactor \left(5x-7\right)\left(x+2\right)Solution StepsSteps Using the Quadratic FormulaSteps Using Direct Factoring MethodView solution stepsSolution Steps 5 x ^ { 2 } + 3 x - 14 Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-14. To find a and b, set up a system to be solved. a+b=3 ab=5\left(-14\right)=-70 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 -70. -1,70 -2,35 -5,14 -7,10 Calculate the sum for each pair. -1+70=69 -2+35=33 -5+14=9 -7+10=3 The solution is the pair that gives sum 3. a=-7 b=10 Rewrite 5x^{2}+3x-14 as \left(5x^{2}-7x\right)+\left(10x-14\right). \left(5x^{2}-7x\right)+\left(10x-14\right) Factor out x in the first and 2 in the second group. x\left(5x-7\right)+2\left(5x-7\right) Factor out common term 5x-7 by using distributive property. \left(5x-7\right)\left(x+2\right) Evaluate \left(5x-7\right)\left(x+2\right)GraphQuizPolynomial5 problems similar to: 5 x ^ { 2 } + 3 x - 14
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2x^2+3x-14http://www.tiger-algebra.com/drill/2x~2_3x-14/ 2x2+3x-14 Final result : (x - 2) • (2x + 7) Step by step solution : Step 1 :Equation at the end of step 1 : (2x2 + 3x) - 14 Step 2 :Trying to factor by splitting the middle term ... How do you solve using the completing the square method \displaystyle{5}{x}^{{2}}+{3}{x}-{1}={0} ?https://socratic.org/questions/how-do-you-solve-using-the-completing-the-square-method-5x-2-3x-1-0 A08 Feb 28, 2016 Given expression is \displaystyle{5}{x}^{{2}}+{3}{x}−{1}={0} We are required to solve for \displaystyle{x} by using *Completing the Square method. Observe that ... x^2+13x-14http://www.tiger-algebra.com/drill/x~2_13x-14/ x2+13x-14 Final result : (x + 14) • (x - 1) Step by step solution : Step 1 :Trying to factor by splitting the middle term 1.1 Factoring x2+13x-14 The first term is, x2 its coefficient ... 5x^2-3x-14http://www.tiger-algebra.com/drill/5x~2-3x-14/ 5x2-3x-14 Final result : (x - 2) • (5x + 7) Step by step solution : Step 1 :Equation at the end of step 1 : (5x2 - 3x) - 14 Step 2 :Trying to factor by splitting the middle term ... x^2+3x-1=0http://www.tiger-algebra.com/drill/X~2_3x-1=0/ x2+3x-1=0 Two solutions were found : x =(-3-√13)/2=-3.303 x =(-3+√13)/2= 0.303 Step by step solution : Step 1 :Trying to factor by splitting the middle term 1.1 Factoring x2+3x-1 ... x^2+3x-108https://www.tiger-algebra.com/drill/x~2_3x-108/ x2+3x-108 Final result : (x + 12) • (x - 9) Step by step solution : Step 1 :Trying to factor by splitting the middle term 1.1 Factoring x2+3x-108 The first term is, x2 its coefficient ...More Items
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CopyCopied to clipboarda+b=3 ab=5\left(-14\right)=-70 Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-14. To find a and b, set up a system to be solved.-1,70 -2,35 -5,14 -7,10 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 -70.-1+70=69 -2+35=33 -5+14=9 -7+10=3 Calculate the sum for each pair.a=-7 b=10 The solution is the pair that gives sum 3.\left(5x^{2}-7x\right)+\left(10x-14\right) Rewrite 5x^{2}+3x-14 as \left(5x^{2}-7x\right)+\left(10x-14\right).x\left(5x-7\right)+2\left(5x-7\right) Factor out x in the first and 2 in the second group.\left(5x-7\right)\left(x+2\right) Factor out common term 5x-7 by using distributive property.5x^{2}+3x-14=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.x=\frac{-3±\sqrt{3^{2}-4\times 5\left(-14\right)}}{2\times 5} 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.x=\frac{-3±\sqrt{9-4\times 5\left(-14\right)}}{2\times 5} Square 3.x=\frac{-3±\sqrt{9-20\left(-14\right)}}{2\times 5} Multiply -4 times 5.x=\frac{-3±\sqrt{9+280}}{2\times 5} Multiply -20 times -14.x=\frac{-3±\sqrt{289}}{2\times 5} Add 9 to 280.x=\frac{-3±17}{2\times 5} Take the square root of 289.x=\frac{-3±17}{10} Multiply 2 times 5.x=\frac{14}{10} Now solve the equation x=\frac{-3±17}{10} when ± is plus. Add -3 to 17.x=\frac{7}{5} Reduce the fraction \frac{14}{10} to lowest terms by extracting and canceling out 2.x=-\frac{20}{10} Now solve the equation x=\frac{-3±17}{10} when ± is minus. Subtract 17 from -3.x=-2 Divide -20 by 10.5x^{2}+3x-14=5\left(x-\frac{7}{5}\right)\left(x-\left(-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{7}{5} for x_{1} and -2 for x_{2}.5x^{2}+3x-14=5\left(x-\frac{7}{5}\right)\left(x+2\right) Simplify all the expressions of the form p-\left(-q\right) to p+q.5x^{2}+3x-14=5\times \frac{5x-7}{5}\left(x+2\right) Subtract \frac{7}{5} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.5x^{2}+3x-14=\left(5x-7\right)\left(x+2\right) Cancel out 5, the greatest common factor in 5 and 5.x ^ 2 +\frac{3}{5}x -\frac{14}{5} = 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 5r + s = -\frac{3}{5} rs = -\frac{14}{5}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 = Cr = -\frac{3}{10} - u s = -\frac{3}{10} + uTwo numbers r and s sum up to -\frac{3}{5} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{5} = -\frac{3}{10}. 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{3}{10} - u) (-\frac{3}{10} + u) = -\frac{14}{5}To solve for unknown quantity u, substitute these in the product equation rs = -\frac{14}{5}\frac{9}{100} - u^2 = -\frac{14}{5}Simplify by expanding (a -b) (a + b) = a^2 – b^2-u^2 = -\frac{14}{5}-\frac{9}{100} = -\frac{289}{100}Simplify the expression by subtracting \frac{9}{100} on both sidesu^2 = \frac{289}{100} u = \pm\sqrt{\frac{289}{100}} = \pm \frac{17}{10} Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable ur =-\frac{3}{10} - \frac{17}{10} = -2 s = -\frac{3}{10} + \frac{17}{10} = 1.400The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.
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Solution StepsSteps Using the Quadratic FormulaSteps Using Direct Factoring MethodView solution stepsSolution Steps 5 x ^ { 2 } + 3 x - 14 Factor the expression by grouping. First, the expression needs to be rewritten as 5x^{2}+ax+bx-14. To find a and b, set up a system to be solved. a+b=3 ab=5\left(-14\right)=-70 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 -70. -1,70 -2,35 -5,14 -7,10 Calculate the sum for each pair. -1+70=69 -2+35=33 -5+14=9 -7+10=3 The solution is the pair that gives sum 3. a=-7 b=10 Rewrite 5x^{2}+3x-14 as \left(5x^{2}-7x\right)+\left(10x-14\right). \left(5x^{2}-7x\right)+\left(10x-14\right) Factor out x in the first and 2 in the second group. x\left(5x-7\right)+2\left(5x-7\right) Factor out common term 5x-7 by using distributive property. \left(5x-7\right)\left(x+2\right) Evaluate \left(5x-7\right)\left(x+2\right)
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