# y=(x^2+5)(3x+1)?

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y = (x^2 + 5) (3x + 1)

Whenever you multiply binomials, remember FOIL.

FOIL = First, Outside, Inside, Last

Whenever you multiply binomials of the form (a + b)(c + d), you would multiply in this fashion:

"First" - Multiply the first terms of each bracket together. In this case, it would be a and c, so a times c is ac.

"Outside" - Multiply the outside terms. This means the first term of the first bracket, and the second term of the second bracket. This should be ad.

"Inside" - Multiply the inside terms, b and c. This will get you bc.

"Last" - Multiply the last terms in each bracket. bd.

It is no different for y = (x^2 + 5) (3x + 1). Let's go through this one at a time:

FIRST: (x^2)(3x) = 3x^3

OUTSIDE: (x^2)(1) = x^2

INSIDE: (5)(3x) = 15x

LAST: (5)(1) = 5

y = 3x^3 + x^2 + 15x + 5.

NOTE: Remember that the minus sign carries over too, when doing FOIL. I'll show you a second example.

y = (2x^2 - 4)(8x + 6)

Doing FOIL:

y = (2x^2)(8x) + (2x^2)(6) - 4(8x) - 4(6)

y = 16x^3 + 12x^2 - 32x - 24

As you can see, when you multiply values out using FOIL, you also have to take into account the minus sign. In this case, we really have a (-4) that we're multiplying.

For more complex ones:

y = (a + b + c) (a + b), you would start with the first value in the first bracket, multiply by everything in the second bracket, go to the second value, multiply everything in the second bracket, and then finally, the same with the third value.

y = (a + b + c) (a + b)

y = (a)(a) + (a)(b) + (b)(a) + (b)(b) + (c)(a) + (c)(b)

y = a^2 + ab + ab + b^2 + ac + bc

y = a^2 + 2ab + b^2 + ac + bc

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• Y=x^2+15

Source(s): sayyed
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• Anonymous

Use the FOIL method:

x^2(3x+1)+5(3x+1)=

3x^3+x^2+15x+5

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• Hi,

If you just want to know what values y could have, you could plug any number in for x and work out the corresponding value of y. There are an infinite number of points you could find.

If you just want to simplify the equation, lots of people already told you how to foil it.

If you are looking for the roots or solutions that would make y = 0, set each factor on the right side equal to 0 and solve. If 3x + 1 = 0, that solves to x = -1/3. If x^2 + 5 = 0, then x^2 = -5 and x = + or - the square root of -5. Since you can't take the square root of a negative number using real numbers, you must go with imaginary numbers and put an "i" in front of the radical, so you get x = + or - "i" radical 5 as your other 2 answers for when y = 0.

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• Assuming the problem was y=((x^2)+5)(3x+1), to find the answer you have to find the zeros of the equation. Where the function hits the graph on the x-axis (y=0) are the solutions to the problem. ((x^2)+5)=0, (3x+1)=0

x^2=-5, x=-1/3

No value of x squared can equal -5, but x=-1/3 works. x=-1/3 is the answer

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y= 3x^3 + 15x + x^2 + 5

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• y = 3x^3 + x^2 + 15x + 5

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• y= 3x^3 + 15x + x^2 + 5

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• y=3x^3+x^2+15x+5

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• = 3x^3 + 1x^2 + 15x + 5

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