# Which of the following is equivalent to 7x^2- 27x - 4? A) (x + 4)(7x - 1)B) (x - 2)(7x + 2)C) (x + 2)(7x - 2)E) x(7x - 27 - 4)

hkj1385 | (Level 1) Assistant Educator

Posted on

7x^2- 27x - 4

This equation is a quadratic equation of type Ax^2 + Bx + C

Now, here A = 7 ; B = -27 & C = -4

A*C = -28

B^2 = 729

Now, -27 = -28 + 1

Thus, the equation can be rewritten as 7x^2 - 28x + x - 4

or, 7x(x-4) +1(x-4)

or, (7x+1)*(x-4)

So, none of the given options are correct.

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to determine which of the given options is equivalent to 7x^2- 27x - 4.

7x^2 - 27x - 4 = 0

=> 7x^2 - 28x + x - 4 = 0

=> 7x(x - 4) + 1(x - 4) = 0

=> (7x + 1)(x - 4)

The options are:

A) (x + 4)(7x - 1) = 7x^2 + 28x - x - 4 = 7x^2 + 27x - 4

B) (x - 2)(7x + 2) = 7x^2 - 14x + 2x - 4 = 7x^2 - 12x - 4

C) (x + 2)(7x - 2) = 7x^2 + 14x - 2x - 4 = 7x^2 + 12x - 4

E) x(7x - 27 - 4) = 7x^2 - 27x - 4x = 7x^2 - 31x

None of the options given are equivalent to the given expression. The factorized form of 7x^2- 27x - 4 = (7x + 1)(x - 4)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

All we have to do is to determine the roots of the quadratic, knowing the fact that the quadratic is the result of the product of two linear factors: (x-x1)(x-x2), where x1 and x2 are the roots of the quadratic.

We'll determine the roots applying quadratic formula:

x1 = [27+sqrt(729+112)]/14

x1=(27+sqrt841)/14

x1=(27+29)/14

x1=4

x2 = -1/7