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(ax+3)(5x^(2)-bx+4)=20x^(3)-9x^(2)-2x+12 The equation above is true for all x, where a and b are constants. What is the value of ab ?

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To solve the equation, we first need to expand the left-hand side of the equation. We multiply (ax+3)(5x^(2)-bx+4) to get ax*5x^(2)-ax*bx+ax*4+3*5x^(2)-3*bx+12 on the left-hand side of the expression.

Afterward, we set the resulting expression equal to the polynomial on the right-hand side, which is 20x^(3)-9x^(2)-2x+12.

Remember that two polynomials are equal for all values of x if and only if their corresponding coefficients are equal.

First of all, let's equate the coefficients for the x^3 terms:
The coefficient from the left side equation is a*5, and on the right side equation, it's 20.
Setting both equal gives us a*5=20.
Solving for a, we find that a=20/5, so a=4.

Next, let's consider the coefficients for the x^2 terms:
On the left side, the coefficient is -ab+15, and on the right, it's -9.
Equating both coefficients gives us -ab+15=-9. If we solve for ab, we obtain ab=15+9, which simplifies to ab=24.

So, the required value of ab is 24.

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