Question

Find all real zeros of number 11

Answer 1

The answer is, $x = 4, \frac{3}{2}, -7$

I'm about to basically type up a report on explaining this answer so bear with me :P

Step 1.) Factor $2x^{3}+ 3x^{2} +65x+84$ using Polynomial Division.

Just a basic rundown of what polynomial division is. When using polynomial division for this problem we find all the factors of our constant, 84, and then see which one, when substituted for x, makes the expression equal zero. After we find our variable that makes the problem equal zero, we divide it into our original expression. Here's how it's done:

These are our factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

strating from one and plugging in these numbers, the first one we come upon that works is 4. So, now we will divide. below is a picture of what it should look like.

After we divide, we get an answer of $( 2x^{2} +11x-21)(x-4)$

Step 2.) split $( 2x^{2} +11x-21)(x-4)$ into two terms

To do that we multiply the coefficient of the first term (2) by the constant (-21)
So: $2*-21=-42$

Now we ask "which two numbers add up to equal 11, but multiply to equal -42". Those two numbers are 14 and -3.

Now we basically split 11x as the sum of 14 and -3. So now our problem should look like this: $(2x^{2}+14x-3x-21) (x-4)$

Step 3.) Factor out the common terms in the first two terms, then the last two like this:

From this: $(2x^{2}+14x-3x-21) (x-4)$

To this: $(2x(x+7)-3(x+7))(x-4)$

Step 4.) Factor the term (x + 7) out of the parenthesis

After we do that, we are left with our problem looking like this:

$(x+7)(2x-3)(x-4)$

The last thing we do is set each parenthesis equal to zero and work it out to find our zeros.

$(x+7) = 0$
$x = -7$

$(2x-3) =0$
$x = \frac{3}{2}$

$(x-4)=0$
$x=4$

That gives us our answer of our zeros being $x = 4, \frac{3}{2}, -7$

Thank you for our question! I hope this helped! Have an amazing day! :D