Respuesta :
The product of 1/2x - 1/4 and 5x^2 - 2x + 6 is 5/2x^3 – 9/4x^2 + 7/2x – 3/2
(a) (1/2x - 1/4) (5x^2 – 2x + 6)
5/2x^3 – 5/4x^2 – 4/4x^2 + 2/4x +6/2x-6/4
(I used 4/4 because it benefits me more when I visually see it)
5/2x^3 -9/4x^2 + 14/4x – 6/4
5/2x^3 – 9/4x^2 + 7/2x – 3/2
(b) No, because with the first part of the problem, the negative goes to ½ instead of 1/4x. When you find the product, instead of finding the exact same answer, you get 5/4x^3 – 3x^2 + 5/2x -3. The numbers when switched, change what they do. 1/4x just multiplies without a negative and ½ does with a negative.
The product of the polynomial [tex](\dfrac{1}{2}x-\dfrac{1}{4})[/tex] and [tex](5x^2-2x+6)[/tex] is [tex]\dfrac{10x^3 -9x^2+ 14x-12}{4}[/tex].
What are polynomial?
Polynomial is an expression that consists of indeterminates(variable) and coefficient, it involves mathematical operations such as addition, subtraction, multiplication, etc, and non-negative integer exponentials.
The product of the polynomial [tex](\dfrac{1}{2}x-\dfrac{1}{4})[/tex] and [tex](5x^2-2x+6)[/tex] can be done in the following way,
[tex](\dfrac{1}{2}x-\dfrac{1}{4})(5x^2-2x+6)\\\\= [(\dfrac{1}{2}x \times 5x^2)- (\dfrac{1}{2}x \times 2x)+(\dfrac{1}{2}x \times 6)] -[(\dfrac{1}{4} \times 5x^2 - (\dfrac{1}{4} \times2x) + (\dfrac{1}{4} \times 6)]\\\\= [\dfrac{5}{4} x^3 - x^2+3x]-[\dfrac{5}{4}x^2 -\dfrac{x}{2}+3]\\\\= \dfrac{5}{4} x^3 - x^2+3x-\dfrac{5}{4}x^2 +\dfrac{x}{2}-3\\\\= \dfrac{5}{2}x^3 -x^2-\dfrac{5}{4}x^2+3x+\dfrac{x}{2}-3\\\\= \dfrac{5}{2}x^3 -\dfrac{4}{4}x^2-\dfrac{5}{4}x^2+\dfrac{6}{2}x+\dfrac{x}{2}-3\\\\[/tex]
[tex]= \dfrac{5}{2}x^3 -\dfrac{9}{4}x^2+\dfrac{7}{2}x-3\\\\= (\dfrac{5 \times 2}{2 \times 2})x^3 -\dfrac{9}{4}x^2+(\dfrac{7 \times 2}{2\times 2})x-\dfrac{3\times 4}{4}\\\\=\dfrac{10x^3}{4} - \dfrac{9x^2}{4} + \dfrac{14x}{4}-\dfrac{12}{4}\\\\=\dfrac{10x^3 -9x^2+ 14x-12}{4}[/tex]
Hence, the product of the polynomial [tex](\dfrac{1}{2}x-\dfrac{1}{4})[/tex] and [tex](5x^2-2x+6)[/tex] is [tex]\dfrac{10x^3 -9x^2+ 14x-12}{4}[/tex].
Learn more about Polynomial:
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