Solve for p, q, and r in 2x^2 − 12x + p = q(x − r)^2 + 10 for all x values

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To solve the equation 2x^2 − 12x + p = q(x − r)^2 + 10 for all x values, the correct values are found by matching coefficients. The expression 2x^2 − 12x can be rewritten as 2(x-3)^2, leading to q = 2 and r = 3. However, the constant term p must be adjusted to 28 to maintain equality across the equation. The key is to ensure that corresponding coefficients on both sides of the equation are equal. Properly expanding and comparing terms clarifies the correct values for p, q, and r.
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2x^2 − 12x + p = q(x − r)^2 + 10 for all values of x find p q and r

i've got

2(x-3)^2 + p = Q(x-r)^2 + 10

so q = 2 r = 3 and p = 10?

now the answers say that Q = 2 r = 3 and p = 28?

can someone explain how to properly go about doing this equation.
 
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2x^2 − 12x is not the same as 2(x-3)^2
 
If a polynomial equation is true for all values of x, then "corresponding coefficients"- that is, coefficients multiplying the same powers of x on opposite sides of the equation- must be the same: multiply the square on the right and compare "corresponding coefficients".
 
kamerling said:
2x^2 − 12x is not the same as 2(x-3)^2



k I've done it now, thx
 

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