Question

Please show me how to solve this question: http://snag.gy/dEc7u.jpg

Answer 1

see if that makes sense first

Answer 2

divisor was \((x+5)\)

which had \(+5\)

so i took took the negative of \(+5\)

that is \(-5\)


then i placed down the coefficients

of the equation

that is

\(5~~21~~0~~-5\)

Answer 3

Oh, I didn't add the zero >-< . I you multiplied the -5 by the 5 21 0 and -5. Does this make it a biomial factor ?

Answer 4

What make the binomial a factor?

Answer 5

i placed down the first coefficient \(5\)

then i multiplied \(-5\) with \(5\)

\(=-25\)

then \(21+(-25)=-4\)

then i multiplied \(-5\) with \(-4\)

\(=20\)

then \(0+20=20\)

then i multiplied \(-5\) with \(20\)

\(=-100\)

then \(-5-100=-105\)

Answer 6

well

\((x+\alpha)\) is a binomial factor of

\(ax^3+bx^2+cx+d\)


if

\(ax^3+bx^2+cx+d=(x+\alpha)(mx^2+nx+v)\)

Answer 7

but our current polynomial is

\(5x^3 + 21x^2 - 5=(x+5)\left(5x^2-4x+20-\dfrac{105}{(x+5)}\right)\)


so as it leaves remainder of \(-\dfrac{105}{(x+5)}\)

it is \(\huge \text{not a binomial factor}\)

Answer 8

I think I got. Thank you very much. I have a similar problem like this one so I will test out what I've learned ^^.