How do you solve $(2 x - \frac{1}{3}) (x + \frac{3}{4}) (x^{2} - 25) = 0$ $(2x - \frac { 1} { 3} ) ( x + \frac { 3} { 4} ) ( x ^ { 2} - 25) = 0$ ?

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Answer 1 · 2017-01-30T23:52:54

Use the Zero Product Property for each of the factors:

Solve $2 x - \frac{1}{3} = 0$ $2x - 1/3=0$ to get $2 x = \frac{1}{3}$ $2x = 1/3$ or $x = \frac{1}{6}$ $x = 1/6$ .

Solve $x + \frac{3}{4} = 0$ $x+3/4=0$ to get $x = - \frac{3}{4}$ $x = -3/4$ .

And last, solve $x^{2} - 25 = 0$ $x^2-25=0$ to get $x^{2} = 25$ $x^2=25$ or $x = \pm 5$ $x=+-5$ .

This factored form of a polynomial equation makes your job much easier!

How do you solve (2x−13)(x+34)(x2−25)=0(2x - \frac { 1} { 3} ) ( x + \frac { 3} { 4} ) ( x ^ { 2} - 25) = 0?