2x^2 + 5x +2 =?

Question

The answer is (2x + 1)( x+ 2) Can someone pls spell how this works? Thanks =)

Answer 1

What you gave was a quadratic equation. You solve this by factoring. First, you must think of pairs of numbers that multiply to 2, the only one being 1 and 2. You also showed that both signs are plus symbols, which means that every number is positive. You then set up the equation as

(2x+ __)(x + __)

because you want to use the FOIL method (first, outer, inner, last) to receive the same original equation. You would have to multiply the 2x by x, then the 2x by either the 2 or 1 (in this case, 2), followed by the second number in the first parenthesis being multiplied by the x and the second number in the second parenthesis. When you are done, your answer looks like

(2x+1)(x+2)

In a quadratic equation, it is always equal to zero, in which it would look like

(2x+1)(x+2)=0.

It may also be a function, in which it may look like

(2x+1)(x+2)=y
or
f(x)=(2x+1)(x+2)

If the example is equal to 0, you can also go further and solve for the value of x. You would set both sets of parenthesis equal to 0, which would look like this:

(2x+1)=0 (x+2)=0

Then you would solve for x and the final answer would be

x={-1/2, -2}

Answer 2

you need two numbers that multiply to 4 (a * c = 2 * 2) and add to 5 (b)
so the answers would be 4 and 1, substitute 4x + 1x in for 5x:

2x^2 + 4x + x + 2 = 0; group the first 2 terms and the last 2 terms:

(2x^2 + 4x) + (x + 2) = 0; factor out the common terms:

2x(x + 2) + (x + 2) = 0; if the terms inside of the brackets are the same you know you are doing it right:

collect the coefficients infront of the brackets and multiply by the bracketed terms:

(2x + 1)(x + 2) = 0

Answer 3

find the multiples of the first coeffiect on the first term which is

1 and 2

then do the same for the 3rd term (2)

1 and 2

now u gotta mix and match them and multiply it out and add them so that the middle term would add up to 5. In this case....2x*2 =4x ....1*x =x x+4x = 5x thus u get the answer right there as (2x + 1)( x+ 2). Now do a quick check. Multipy it out u get....2x^2 + 5x +2. THERE!! next problem!

Answer 4

Well you can use the FOIL method (First Outside Inside Last) to check the answer so

2x * x + 2x * 2 + 1 * x + 1 * 2

giving you

2x^2 + 5x + 2

When you factor it out, it allows you to find the zeroes of the function much easier by setting each factor to 0.

Answer 5

look across the problem 2x times x gives u the 2xsquared and 2 times 1 gives u 2 then 2x times 2 is 4 plus 1 gives u 5 in the middle

Answer 6

1) split the quadratic into 4 terms.
(2x^2+4x)+(x+2)=0

2) factor out 2x from the first set of parentheses.
2x(x+2)+(1)(x+2)=0

3) (2x+1)(x+2)=0

i can't remember the name of the method i used, but i goes like this: ax(x+n)+b(x+n)=(ax+b)(x+n) where x is a variable and a and b are any real numbers.

Answer 7

U know FOIL?
Times the first terms, then the outside, then the inside, then the last.

To figure the solution out see wat binomials when FOILed work out to 2x^2+5x+2+?

U know the first term has to be mulitplied by 2x and x(thats the only two that work) to get 2x^2.
Then u do that 4 the rest.

Or ask ur math teacher:)

Answer 8

(2x+1)(x+2)
2x*x=2x^2
1*x=x
2x*2=4x
1*2=2
2x^2+x+4x+2
2x^2+5x+2