Question

Find the value of the polynomial 3xy+7x2−3x2y+3y2x−2x2−2xy+4x2y−2y2x, when x=1 and y=−2

Answer 1

solving the polynomial $3xy+7x^2-3x^2y+3y^2x-2x^2-2xy+4x^2y-2y^2x\,\,when\,\,x=1\,\,and\,\,y=-2$ we get the value -16

Step-by-step explanation:

We need to find the value of polynomial $3xy+7x^2-3x^2y+3y^2x-2x^2-2xy+4x^2y-2y^2x\,\,when\,\,x=1\,\,and\,\,y=-2$

Putting value of x as 1 and y=-2 and finding the value of polynomial.

$3xy+7x^2-3x^2y+3y^2x-2x^2-2xy+4x^2y-2y^2x\\=3(1)(-2)+7(1)^2(-2)-3(1)^2(-2)+3(-2)^2(1)-2(1)^2-2(1)(-2)+4(1)^2(-2)-2(-2)^2(1)\\=3(-2)+7(-2)-3(-2)+3(4)-2+4-8-2(4)\\=-6-14+6+12-2+4-8-8\\=-16$

So, solving the polynomial $3xy+7x^2-3x^2y+3y^2x-2x^2-2xy+4x^2y-2y^2x\,\,when\,\,x=1\,\,and\,\,y=-2$ we get the value -16

Keywords: Solving polynomials

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