Question

Factor the trinomial below.

x2 - 14x + 45

Answer 1

`x_(1)=9, x_2=5`

Explanation:

Recall a quadratic formula for quadratic equation
`ax^2+bx+c=0`:
`x_(1/2)=(-b\pm √(b^2 -4ac))/(2a)`

Notice the constant in the given equation
`x^2-14x+45`:

`a=1, b=-14, c=45`

Substitute the values into the formula:

`x_(1/2)=(-(-14)\pm √((-14)^2 -4\cdot 1\cdot 45))/(2\cdot 1)`

Calculate the values:

`x_(1/2)=(14\pm √(196 -180))/(2)=(14\pm√(16))/(2)=(14\pm 4)/(2)`

It follows that there are two solutions:

`x_1=(14+4)/(2)=(18)/(2)=9` and
`x_2=(14-4)/(2)=(10)/(2)=5`.

Answer 2

(x-9) (x-5)

x = 9 x = 5

Explanation:

find out which 2 numbers when multiplied give 45 and

when added give -14

they are -9 and -5

(x-9) (x-5)

x = 9 x = 5