Solve 7x-3-4x^2

$$7x-3-4 { x }^{ 2 }$$

Factor

$$\left(1-x\right)\left(4x-3\right)$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-4x^{2}+7x-3$$

Factor the expression by grouping. First, the expression needs to be rewritten as -4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=7 ab=-4\left(-3\right)=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 7.

$$a=4 b=3$$

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

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

Factor out 4x in the first and -3 in the second group.

$$4x\left(-x+1\right)-3\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(4x-3\right)$$

Steps Using the Quadratic Formula

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.

$$-4x^{2}+7x-3=0$$

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.

$$x=\frac{-7±\sqrt{7^{2}-4\left(-4\right)\left(-3\right)}}{2\left(-4\right)}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\left(-4\right)\left(-3\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-7±\sqrt{49+16\left(-3\right)}}{2\left(-4\right)}$$

Multiply 16 times -3.

$$x=\frac{-7±\sqrt{49-48}}{2\left(-4\right)}$$

Add 49 to -48.

$$x=\frac{-7±\sqrt{1}}{2\left(-4\right)}$$

Take the square root of 1.

$$x=\frac{-7±1}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-7±1}{-8}$$

Now solve the equation x=\frac{-7±1}{-8} when ± is plus. Add -7 to 1.

$$x=-\frac{6}{-8}$$

Reduce the fraction \frac{-6}{-8} to lowest terms by extracting and canceling out 2.

$$x=\frac{3}{4}$$

Now solve the equation x=\frac{-7±1}{-8} when ± is minus. Subtract 1 from -7.

$$x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x=1$$

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

$$-4x^{2}+7x-3=-4\left(x-\frac{3}{4}\right)\left(x-1\right)$$

Subtract \frac{3}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+7x-3=-4\times \frac{-4x+3}{-4}\left(x-1\right)$$

Cancel out 4, the greatest common factor in -4 and 4.

$$-4x^{2}+7x-3=\left(-4x+3\right)\left(x-1\right)$$

Evaluate

$$\left(1-x\right)\left(4x-3\right)$$