By factor theorem, substituting x = 2 and x = 1/2 both make the polynomial equal 0. Equating the resulting expressions 4p + 10 + r = 0 and p/4 + 5/2 + r = 0 simplifies perfectly to prove ...
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We divide the given quadratic area expression by the given width binomial. Factoring the numerator 2x square + 7x + 3 yields (2x + 1)(x + 3) and cancelling the common width factor leaves the length.
: Setting up the equation as x + 1/x = 10/3 creates a quadratic form 3x square – 10x + 3 = 0. Factoring this gives (3x – 1)(x – 3) = 0, which provides the two possible numerical answers.
A polynomial of degree 1 can be 3x plus 2. A polynomial of degree 2 can be x square plus 5x plus 1. A polynomial of degree 3 can be 2x cube minus x plus 4.
If the length of the rectangle is x cm, then the width becomes 10 minus x cm because the total perimeter is 20 cm. Therefore, the area of the rectangle is x multiplied by 10 minus x.