Solve 4x^2-35x+49

$$4 { x }^{ 2 } -35x+49$$

Factor

$$\left(x-7\right)\left(4x-7\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 4x^{2}+ax+bx+49. To find a and b, set up a system to be solved.

$$a+b=-35 ab=4\times 49=196$$

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 196.

$$-1,-196 -2,-98 -4,-49 -7,-28 -14,-14$$

Calculate the sum for each pair.

$$-1-196=-197 -2-98=-100 -4-49=-53 -7-28=-35 -14-14=-28$$

The solution is the pair that gives sum -35.

$$a=-28 b=-7$$

Rewrite 4x^{2}-35x+49 as \left(4x^{2}-28x\right)+\left(-7x+49\right).

$$\left(4x^{2}-28x\right)+\left(-7x+49\right)$$

Factor out 4x in the first and -7 in the second group.

$$4x\left(x-7\right)-7\left(x-7\right)$$

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

$$\left(x-7\right)\left(4x-7\right)$$

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.

$$4x^{2}-35x+49=0$$

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.

$$x=\frac{-\left(-35\right)±\sqrt{\left(-35\right)^{2}-4\times 4\times 49}}{2\times 4}$$

Square -35.

$$x=\frac{-\left(-35\right)±\sqrt{1225-4\times 4\times 49}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-35\right)±\sqrt{1225-16\times 49}}{2\times 4}$$

Multiply -16 times 49.

$$x=\frac{-\left(-35\right)±\sqrt{1225-784}}{2\times 4}$$

Add 1225 to -784.

$$x=\frac{-\left(-35\right)±\sqrt{441}}{2\times 4}$$

Take the square root of 441.

$$x=\frac{-\left(-35\right)±21}{2\times 4}$$

The opposite of -35 is 35.

$$x=\frac{35±21}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{35±21}{8}$$

Now solve the equation x=\frac{35±21}{8} when ± is plus. Add 35 to 21.

$$x=\frac{56}{8}$$

Divide 56 by 8.

$$x=7$$

Now solve the equation x=\frac{35±21}{8} when ± is minus. Subtract 21 from 35.

$$x=\frac{14}{8}$$

Reduce the fraction \frac{14}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{7}{4}$$

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 \frac{7}{4} for x_{2}.

$$4x^{2}-35x+49=4\left(x-7\right)\left(x-\frac{7}{4}\right)$$

Subtract \frac{7}{4} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$4x^{2}-35x+49=4\left(x-7\right)\times \frac{4x-7}{4}$$

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

$$4x^{2}-35x+49=\left(x-7\right)\left(4x-7\right)$$

Evaluate

$$\left(x-7\right)\left(4x-7\right)$$