Question

What is the remainder of (4x2 + 7x-1)= (4 + x)?
A. -9x – 1
B.23x – 1
C.35
D.-37​

Answer 1

Answer: I don't know what you meant by remainder but i hope this helps :)

`x=(-3+√(29))/(4),\:x=-(3+√(29))/(4)\\`

Explanation:

`\left(4x^2+7x-1\right)=\left(4+x\right)\\\mathrm{Refine}\\4x^2+7x-1=4+x\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\4x^2+7x-1-x=4+x-x\\Simplify\\4x^2+6x-1=4\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\4x^2+6x-1-4=4-4\\\mathrm{Simplify}\\4x^2+6x-5=0\\\mathrm{For\:}\quad a=4,\:b=6,\:c=-5:\\\quad x_(1,\:2)=(-6\pm √(6^2-4* \:4\left(-5\right)))/(2* \:4)`

`(-6+√(6^2-4* \:4\left(-5\right)))/(2* \:4)\\=(-6+√(6^2+4* \:4* \:5))/(2* \:4)\\=(-6+√(116))/(2* \:4)\\=(-6+√(116))/(8)\\\\Let\: simplify\: ; -6+2√(29)\\=-2* \:3+2√(29)\\=2\left(-3+√(29)\right)\\=(2\left(-3+√(29)\right))/(8)\\=(-3+√(29))/(4)\\`

`(-6-√(6^2-4* \:4\left(-5\right)))/(2* \:4)\\\\=(-6-√(6^2+4* \:4* \:5))/(2* \:4)\\\\=(-6-√(116))/(2* \:4)\\\\=(-6-2√(29))/(8)\\\\=-(2\left(3+√(29)\right))/(8)\\\\=-(3+√(29))/(4)\\\\\\x=(-3+√(29))/(4),\:x=-(3+√(29))/(4)`

Answer 2

C. 35

Explanation:

The synthetic division is shown below. The remainder is the number at lower right of the array, 35.

The remainder from division by x+4 is also the value of the quadratic evaluated at x=-4:

4x² +7x -1 = (4x +7)x -1

= (4(-4) +7)(-4) -1 = (-16 +7)(-4) -1 = 36 -1 = 35