Question

F(x)=x^2-14x+33 enter the quadratic function in factored form

Answer 1

Answer:

$F(x)=(x-11)(x-3)$

Step-by-step explanation:

we have

$F(x)=x^{2} -14x+33$

Find the zeros of the function

F(x)=0

$0=x^{2} -14x+33$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$-33=x^{2} -14x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$-33+49=x^{2} -14x+49$

$16=x^{2} -14x+49$

Rewrite as perfect squares

$16=(x-7)^{2}$

square root both sides

$(x-7)=(+/-)4$

$x=(+/-)4+7$

$x=(+)4+7=11$

$x=(-)4+7=3$

so

The factors are

(x-11) and (x-3)

therefore

$F(x)=(x-11)(x-3)$