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Solve 2x(3x+5)+3(3x+5)=ax 2 +bx+c

User Baub
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2 Answers

5 votes

We can start by simplifying the left side of the equation using the distributive property:

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

= 6x^2 + 10x + 9x + 15

= 6x^2 + 19x + 15

Now we can compare this expression with the right side of the equation, which is a polynomial in x with unknown coefficients a, b, and c:

ax^2 + bx + c

Since the two sides are equal, their corresponding coefficients must be equal as well. This gives us a system of three equations in three unknowns:

a = 6 (the coefficient of x^2)

b = 19 (the coefficient of x)

c = 15 (the constant term)

Therefore, the solution to the equation 2x(3x+5)+3(3x+5)=ax^2+bx+c is:

2x(3x+5)+3(3x+5) = 6x^2 + 19x + 15

User Brown Smith
by
8.0k points
6 votes

Answer:

6x² + 19x + 15

Explanation:

2x(3x+5)+3(3x+5)

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

= 6x² + 9x + 10x + 15

= 6x² + 19x + 15

So, the answer is 6x² + 19x + 15

User SmellyCat
by
7.6k points

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