Question

f(x)=2x^2+7x+3

Use grouping method

Use quadratic method

Use completing the square method

Answer 1

Do you know how to do any of these?

Answer 2

No this is a pre-test @whpalmer4

Answer 3

But can you help me?

Answer 4

@Hero @whpalmer4 ?

Answer 5

As an example,

Suppose

f(x) = x^2 + 5x + 6

Answer 6

Then, to factor by grouping, find two numbers m and n so that

m × n = 6
m + n = 5

Notice that 2 × 3 = 6 and 2 + 3 = 5

Basically, now, since we have found the values of m and n, then substitute 5 with 2 + 3:
x^2 + 5x + 6 = x^2 + (2 + 3)x + 6
= x^2 + 2x + 3x + 6


Now factor the first two terms, then factor the last two terms.

= x(x + 2) + 3(x + 2)

Now factor the remaining term (x + 2) that is common to both factorizations:
= (x + 2)(x + 3)

So x^2 + 5x + 6 = (x + 2)(x + 3)

Answer 7

To factor by quadratic equation \(x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Let a = 1, b = 5 and c = 6
Then

\(x = \dfrac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)}\)

\(x = \dfrac{-5 \pm \sqrt{25 - 24}}{2}\)

\(x = \dfrac{-5 \pm \sqrt{1}}{2}\)

\(x = \dfrac{-5 \pm 1}{2}\)

\(x = \dfrac{\pm 1 - 5}{2}\)

\(x = \pm \dfrac{1 - 5}{2}\)

\(x = - \dfrac{1 + 5}{2} \text{or} \dfrac{1 - 5}{2}\)

\(x = -\dfrac{6}{2} \text{or} \dfrac{-4}{2}\)

\(x = -3 \space \text{or} -2\)

Answer 8

Then

x = -3
x + 3 = 0

or

x = -2
x + 2 = 0

and

(x + 3)(x + 2) = 0

by zero product property

Answer 9

For complete the square:

Let f(x) = 0

Then x^2 + 5x + 6 = 0

Subtract 6 from both sides:
x^2 + 5x = -6

Then add \(\left(\dfrac{b}{2}\right)^2\) to both sides

Since b = 5, then \(\left(\dfrac{b}{2}\right)^2 = \dfrac{5^2}{4} = \dfrac{25}{4}\)

So add \(\dfrac{25}{4}\) to both sides:

\(x^2 + 5x + \dfrac{25}{4} = \dfrac{25}{4} - 6\)

\(x^2 + 5x + \dfrac{25}{4} = \dfrac{25}{4} - \dfrac{24}{4}\)

\(x^2 + 5x + \dfrac{25}{4} = \dfrac{1}{4}\)

Remember that in order to express the right side as a binomial square we need
mn = 25/4
m + n = 5

In this case
m = n = 5/2

Since
(5/2)(5/2) = 25/4
and 5/2 + 5/2 = 10/2 = 5

So

\(x^2 + 5x + \dfrac{25}{4} = \left(x + \dfrac{5}{2}\right)^2\)

and

\(\left(x + \dfrac{5}{2}\right)^2 = \dfrac{1}{4}\)

Take the square root of both sides:
\(x + \dfrac{5}{2} = \pm\dfrac{1}{2}\)

Isolate x:
\(x = \pm\dfrac{1}{2} - \dfrac{5}{2}\)

\(x = -2,\)
\(x = -3\)

\(x + 2 = 0\)
\(x + 3 = 0\)
\((x + 2)(x + 3) = 0\)
\((x + 2)(x + 3) = f(x)\)

Answer 10

Man thank you so much @hero helps so much omg

Answer 11

yw