Type a math problem

This site uses cookies for analytics, personalized content and ads. By continuing to browse this site, you agree to this use. Learn more

Type a math problem

Factor

\left(x-4\right)\left(x-3\right)

$(x−4)(x−3)$

Solution Steps

Steps Using the Quadratic Formula

Steps Using Direct Factoring Method

Solution Steps

x^2-7x+12

$x_{2}−7x+12$

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

Factor the expression by grouping. First, the expression needs to be rewritten as $x_{2}+ax+bx+12$. To find $a$ and $b$, set up a system to be solved.

a+b=-7 ab=1\times 12=12

$a+b=−7$ $ab=1×12=12$

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 12.

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 $12$.

-1,-12 -2,-6 -3,-4

$−1,−12$ $−2,−6$ $−3,−4$

Calculate the sum for each pair.

Calculate the sum for each pair.

-1-12=-13 -2-6=-8 -3-4=-7

$−1−12=−13$ $−2−6=−8$ $−3−4=−7$

The solution is the pair that gives sum -7.

The solution is the pair that gives sum $−7$.

a=-4 b=-3

$a=−4$ $b=−3$

Rewrite x^{2}-7x+12 as \left(x^{2}-4x\right)+\left(-3x+12\right).

Rewrite $x_{2}−7x+12$ as $(x_{2}−4x)+(−3x+12)$.

\left(x^{2}-4x\right)+\left(-3x+12\right)

$(x_{2}−4x)+(−3x+12)$

Factor out x in the first and -3 in the second group.

Factor out $x$ in the first and $−3$ in the second group.

x\left(x-4\right)-3\left(x-4\right)

$x(x−4)−3(x−4)$

Factor out common term x-4 by using distributive property.

Factor out common term $x−4$ by using distributive property.

\left(x-4\right)\left(x-3\right)

$(x−4)(x−3)$

Evaluate

\left(x-4\right)\left(x-3\right)

$(x−4)(x−3)$

Graph

Share

Copy

Copied to clipboard

a+b=-7 ab=1\times 12=12

Factor the expression by grouping. First, the expression needs to be rewritten as x^{2}+ax+bx+12. To find a and b, set up a system to be solved.

-1,-12 -2,-6 -3,-4

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 12.

-1-12=-13 -2-6=-8 -3-4=-7

Calculate the sum for each pair.

a=-4 b=-3

The solution is the pair that gives sum -7.

\left(x^{2}-4x\right)+\left(-3x+12\right)

Rewrite x^{2}-7x+12 as \left(x^{2}-4x\right)+\left(-3x+12\right).

x\left(x-4\right)-3\left(x-4\right)

Factor out x in the first and -3 in the second group.

\left(x-4\right)\left(x-3\right)

Factor out common term x-4 by using distributive property.

x^{2}-7x+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.

x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{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{-\left(-7\right)±\sqrt{49-4\times 12}}{2}

Square -7.

x=\frac{-\left(-7\right)±\sqrt{49-48}}{2}

Multiply -4 times 12.

x=\frac{-\left(-7\right)±\sqrt{1}}{2}

Add 49 to -48.

x=\frac{-\left(-7\right)±1}{2}

Take the square root of 1.

x=\frac{7±1}{2}

The opposite of -7 is 7.

x=\frac{8}{2}

Now solve the equation x=\frac{7±1}{2} when ± is plus. Add 7 to 1.

x=4

Divide 8 by 2.

x=\frac{6}{2}

Now solve the equation x=\frac{7±1}{2} when ± is minus. Subtract 1 from 7.

x=3

Divide 6 by 2.

x^{2}-7x+12=\left(x-4\right)\left(x-3\right)

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute 4 for x_{1} and 3 for x_{2}.

x ^ 2 -7x +12 = 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 = 7 rs = 12

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{7}{2} - u s = \frac{7}{2} + u

Two numbers r and s sum up to 7 exactly when the average of the two numbers is \frac{1}{2}*7 = \frac{7}{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{7}{2} - u) (\frac{7}{2} + u) = 12

To solve for unknown quantity u, substitute these in the product equation rs = 12

\frac{49}{4} - u^2 = 12

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

-u^2 = 12-\frac{49}{4} = -\frac{1}{4}

Simplify the expression by subtracting \frac{49}{4} on both sides

u^2 = \frac{1}{4} u = \pm\sqrt{\frac{1}{4}} = \pm \frac{1}{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{7}{2} - \frac{1}{2} = 3 s = \frac{7}{2} + \frac{1}{2} = 4

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

Back to top