Question

\large x^2-x-90 or \large x^2-1x-90

a
(x - 9)(x + 10)
b
(x - 9)(x - 10)
c
(x + 9)(x + 10)
d
(x + 9)(x - 10)

Answer 1

To factor the quadratic expression \( x^2 - x - 90 \), we need to find two numbers that multiply to \(-90\) (the constant term) and add up to \(-1\) (the coefficient of \(x\)).

Let's look for pairs of factors of \(-90\):

- \(1\) and \(-90\) (sum: -89)
- \(-1\) and \(90\) (sum: 89)
- \(2\) and \(-45\) (sum: -43)
- \(-2\) and \(45\) (sum: 43)
- \(3\) and \(-30\) (sum: -27)
- \(-3\) and \(30\) (sum: 27)
- \(5\) and \(-18\) (sum: -13)
- \(-5\) and \(18\) (sum: 13)
- \(6\) and \(-15\) (sum: -9)
- \(-6\) and \(15\) (sum: 9)
- \(9\) and \(-10\) (sum: -1)
- \(-9\) and \(10\) (sum: 1)

The correct pair is \(9\) and \(-10\) since:

\[
9 + (-10) = -1 \quad \text{and} \quad 9 \cdot (-10) = -90
\]

Thus, we can write:

\[
x^2 - x - 90 = (x - 10)(x + 9)
\]

So the correct factorization is:

\[
(x + 9)(x - 10)
\]

Looking at your options, the correct answer is:

**d** \((x + 9)(x - 10)\)