Factor

\left(x-7\right)\left(x+10\right)

Evaluate

\left(x-7\right)\left(x+10\right)

Graph

Quiz

Polynomial5 problems similar to: x ^ { 2 } + 3 x - 70## Similar Problems from Web Search

4x^2+3x-7

http://www.tiger-algebra.com/drill/4x~2_3x-7/

2x^2+3x-4

2x2+3x-4Final result : 2x2 + 3x - 4 Step by step solution : Step 1 :Equation at the end of step 1 : (2x2 + 3x) - 4 Step 2 :Trying to factor by splitting the middle term 2.1Factoring ...

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a+b=3 ab=1\left(-70\right)=-70

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx-70. 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(x^{2}-7x\right)+\left(10x-70\right)

Rewrite x^{2}+3x-70 as \left(x^{2}-7x\right)+\left(10x-70\right).

x\left(x-7\right)+10\left(x-7\right)

Factor out x in the first and 10 in the second group.

\left(x-7\right)\left(x+10\right)

Factor out common term x-7 by using distributive property.

x^{2}+3x-70=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\left(-70\right)}}{2}

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\left(-70\right)}}{2}

Square 3.

x=\frac{-3±\sqrt{9+280}}{2}

Multiply -4 times -70.

x=\frac{-3±\sqrt{289}}{2}

Add 9 to 280.

x=\frac{-3±17}{2}

Take the square root of 289.

x=\frac{14}{2}

Now solve the equation x=\frac{-3±17}{2} when ± is plus. Add -3 to 17.

x=7

Divide 14 by 2.

x=-\frac{20}{2}

Now solve the equation x=\frac{-3±17}{2} when ± is minus. Subtract 17 from -3.

x=-10

Divide -20 by 2.

x^{2}+3x-70=\left(x-7\right)\left(x-\left(-10\right)\right)

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 7 for x_{1} and -10 for x_{2}.

x^{2}+3x-70=\left(x-7\right)\left(x+10\right)

Simplify all the expressions of the form p-\left(-q\right) to p+q.

x ^ 2 +3x -70 = 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.

r + s = -3 rs = -70

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{3}{2} - u s = -\frac{3}{2} + u

Two numbers r and s sum up to -3 exactly when the average of the two numbers is \frac{1}{2}*-3 = -\frac{3}{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{3}{2} - u) (-\frac{3}{2} + u) = -70

To solve for unknown quantity u, substitute these in the product equation rs = -70

\frac{9}{4} - u^2 = -70

Simplify by expanding (a -b) (a + b) = a^2 – b^2

-u^2 = -70-\frac{9}{4} = -\frac{289}{4}

Simplify the expression by subtracting \frac{9}{4} on both sides

u^2 = \frac{289}{4} u = \pm\sqrt{\frac{289}{4}} = \pm \frac{17}{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{3}{2} - \frac{17}{2} = -10 s = -\frac{3}{2} + \frac{17}{2} = 7

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.