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Ax2+bx+c=(5x−6)(2x−1)

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Answer:

To solve the equation Ax^2 + bx + c = (5x - 6)(2x - 1), we need to expand the right side of the equation using the distributive property.

(5x - 6)(2x - 1) can be expanded as follows:

(5x)(2x) + (5x)(-1) + (-6)(2x) + (-6)(-1)

Which simplifies to:

10x^2 - 5x - 12x + 6

Combining like terms, we get:

10x^2 - 17x + 6

So, the expanded form of (5x - 6)(2x - 1) is 10x^2 - 17x + 6.

Now, we can equate this expression to the left side of the equation:

Ax^2 + bx + c = 10x^2 - 17x + 6

By comparing the coefficients of corresponding powers of x on both sides, we can determine the values of A, b, and c.

Comparing the coefficient of x^2:

A = 10

Comparing the coefficient of x:

b = -17

Comparing the constant terms:

c = 6

Therefore, the values of A, b, and c are A = 10, b = -17, and c = 6.

In summary, the values of A, b, and c for the equation Ax^2 + bx + c = (5x - 6)(2x - 1) are A = 10, b = -17, and c = 6.

User Nbevans
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