Solve the given equation by factoring. 2x2+7x+6=02x^2+7x+6=02x2+7x+6=0

x = − 2 , x = − 3 2 x=-2,x=-\frac32 x = − 2 , x = − 2 3

Solve the given equation by factoring. 2 x 2 + 7 x + 6 = 0 2x^2+7x+6=0 2 x 2 + 7 x + 6 = 0

Hey everyone. We want to solve this quadratic equation by factoring and here I have 2x squared plus 7x plus 6 is equal to 0. Now step number 1 is to write this quadratic equation in standard form with all my terms on the same side in descending order of power and it looks like they already are in standard form so step 1 is done. Now step 2 is going to be to factor this completely. So looking at the quadratic equation I have, I know that there are 3 terms and there's not a factoring formula that I think will work here so I'm going to go ahead and use the a c method to factor this. So I want to find two numbers that multiply to a c and add to b. So in this case, a is 2, b is 7, and c is 6. So if I have a times c, that's gonna be 2 times 6, which is 12. So I wanna find 2 numbers that multiply to 12 and then add to b which is 7. So let's go ahead and think about some numbers there. Well, I know that 3 and 4 multiply to 12 and that seems like something that could add to 7 so let's try that. 3 times 4 is 12 and 3+4 is equal to 7. So 3 and 4 are going to be our factors here. Now since we're dealing with a problem where a is not equal to 1 that means that I need to go ahead and write this as 2x squared plus 3x plus 4x plus 6 where 3 and 4 were those factors that I just found. And now we can go ahead and factor by grouping. So So I'm just gonna group my first two terms and my last two terms in order to do that. 2x squared is really just 2 times x times x. 3x is just 3 times x. 4x is 2 times 2 times x, and 6 is just 2 times times 3. So looking at my first group, what common factor do I have there? Well, I have an x in both terms, so I'm gonna go ahead and pull out an x here. If I pull out an x of those first two terms, I will be left with 2x plus 3. So that's my first group. Let's take a look at our second group here. So in my second group, both of these have a 2. So I'm gonna go ahead and pull a 2 out in the front, and then I have left over 2 x plus 3. So it worked out perfectly. I have these two factors here that are the same. So this really is just x plus 2 taking out that times 2x plus 3. So now that I have factored this completely, I can move on to step 3, which is to set each of my factors equal to 0. So starting with this x+2, if I set x plus 2 equals 0, and then 2x plus 3 equal to 0 as well, and let's solve. So I can just subtract my 2 here, canceling it, leaving me with x is equal to negative 2. And then over here, I can subtract 3, leaving me with 2x equals negative 3, but I have one more step here which is to divide by 2 leaving me with x is equal to negative three halves. So my two solutions here are x equals a negative 2 and x equals negative three halves and I've completed step 3. Now we could also go ahead and plug those back in to make sure that these are correct, which I will leave up to you. See you in the next video.

Solve the given equation by factoring. 2 x 2 + 7 x + 6 = 0 2x^2+7x+6=0 2 x 2 + 7 x + 6 = 0

x = − 2 , x = − 3 2 x=-2,x=-\frac32 x = − 2 , x = − 2 3

x = 3 , x = 6 7 x=3,x=\frac67 x = 3 , x = 7 6

x = − 2 , x = 0 x=-2,x=0 x = − 2 , x = 0

x = 2 , x = 3 2 x=2,x=\frac32 x = 2 , x = 2 3

1. Equations and Inequalities

Intro to Quadratic Equations

Precalculus

1. Equations and inequalities

Intro to Quadratic Equations

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Multiple Choice

Solve the given equation by factoring. $2x^2+7x+6=0$

$x=3,x=\frac67$

$x=-2,x=0$

$x=2,x=\frac32$

$x=-2,x=-\frac32$

Verified step by step guidance

Start by writing the quadratic equation in standard form: $2x^2 + 7x + 6 = 0$.

To factor the quadratic equation, look for two numbers that multiply to $2 \times 6 = 12$ and add to $7$. These numbers are $3$ and $4$.

Rewrite the middle term $7x$ using the numbers found: $2x^2 + 3x + 4x + 6 = 0$.

Group the terms to factor by grouping: $(2x^2 + 3x) + (4x + 6) = 0$.

Factor out the greatest common factor from each group: $x(2x + 3) + 2(2x + 3) = 0$. Now, factor out the common binomial factor: $(x + 2)(2x + 3) = 0$.

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