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The question is incomplete in the sense that there is no +/- sign between q2 and 2q.

Considering plus '+' sign.

q3 - q2 + 2q - 2

q2(q - 1) + 2(q - 1)

(q2 + 2)(q - 1)

Considering minus '-' sign.

q3 - q2 - 2q - 2

The expression can't be factored.

However if we consider the question complete then it is written as:

q3 - q2(2q) - 2

q3 - 2q3 - 2

-q3 - 2

-(q + 21/3)(q2 - 21/3 + 22/3)

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Related Questions

What is the perpendicular bisector equation of the line segment of p q and 7p 3q?

In its general form it works out as: 3px+qy-12p2-2q2 = 0Improved Answer:-Points:(p, q) and (7p, 3q)Slope: q/3pPerpendicular slope: -3p/qMidpoint: (4p, 2q)Equation: y-2q = -3p/q(x-4p) => yq = -3px+12p2+2q2Perpendicular bisector equation in its general form: 3px+qy-12x2-2q2 = 0


How do you work out and find the perpendicular bisector equation meeting the straight line segment of p q and 7p 3q?

First find the mid-point of the line segment which will be the point of intersection of the perpendicular bisector. Then find the slope or gradient of the line segment whose negative reciprocal will be the perpendicular bisector's slope or gradient. Then use y -y1 = m(x -x1) to find the equation of the perpendicular bisector. Mid-point: (7p+p)/2 and (3q+q)/2 = (4p, 2q) Slope or gradient: 3q-q/7p-p = 2q/6p = q/3p Slope of perpendicular bisector: -3p/q Equation: y -2q = -3p/q(x -4p) y = -3px/q+12p2/q+2q Multiply all terms by q to eliminate the fractions: qy = -3px+12p2+2q2 Which can be expressed in the form of: 3px+qy-12p2-2q2 = 0


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How do you form an equation for the perpendicular bisector of the line segment joining the points of p q and 7p 3q showing all details of your work?

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What is the quadratic 30q3 plus 14q2-4q equals 0?

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(3)(q-11) = 3q-33


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What is the perpendicular bisector equation of a line segment with endpoints of p q and 7p 3q?

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