How to find remainder

M

mimie

New member
Joined
May 21, 2019
Messages
42
When a polynomial p(x) of degree n≥2 has a remainder of 7 when it is divided by (2x−3) and a remainder of −2 when it is divided by (4x+3). Find the remainder of p(x) when it is divided by (8x^2 −6x−9).

I don't know to do this question, please help me. Thanks.
 
In symbols:

[math]\dfrac{p(x)}{2x-3} = q(x) + \dfrac{7}{2x-3}\;for\;some\;q(x).[/math]
[math]\dfrac{p(x)}{4x+3} = r(x) + \dfrac{-2}{4x+3}\;for\;some\;r(x).[/math]
and we seek N, such that:

[math]\dfrac{p(x)}{(2x-3)(4x+3)} = s(x) + \dfrac{N}{(2x-3)(4x+3)}\;for\;some\;s(x)\;and\;N.[/math]
Does the visual help to understand what is asked?

Did you notice that [math](2x-3)(4x+3) = 8x^{2} - 6x - 9[/math]?

Is N a Real Number or should it be N(x) of degree possibly one (1)?

What can we do symbolically if we promise that [math]x \ne 3/2\;and\;x\ne -3/4[/math]?
 
tkhunny said:
In symbols:

[math]\dfrac{p(x)}{2x-3} = q(x) + \dfrac{7}{2x-3}\;for\;some\;q(x).[/math]
[math]\dfrac{p(x)}{4x+3} = r(x) + \dfrac{-2}{4x+3}\;for\;some\;r(x).[/math]
and we seek N, such that:

[math]\dfrac{p(x)}{(2x-3)(4x+3)} = s(x) + \dfrac{N}{(2x-3)(4x+3)}\;for\;some\;s(x)\;and\;N.[/math]
Does the visual help to understand what is asked?

Did you notice that [math](2x-3)(4x+3) = 8x^{2} - 6x - 9[/math]?

Is N a Real Number or should it be N(x) of degree possibly one (1)?

What can we do symbolically if we promise that [math]x \ne 3/2\;and\;x\ne -3/4[/math]?
Click to expand...


I think N is degree one, still not sure. But I have no idea how to proceed to get the N .
 
Address the last question, first. Make THREE expressions for p(x).
 
mimie said:
When a polynomial p(x) of degree n≥2 has a remainder of 7 when it is divided by (2x−3) and a remainder of −2 when it is divided by (4x+3). Find the remainder of p(x) when it is divided by (8x^2 −6x−9).

I don't know to do this question, please help me. Thanks.
Click to expand...
Let [MATH]a(x)[/MATH] and [MATH]b(x)[/MATH] be the quotients obtained by dividing [MATH]p(x)[/MATH] by [MATH]2x-3[/MATH] and by [MATH]4x+3[/MATH], respectively. Then, by the division algorithm,

[MATH](2x-3)a(x)+7=p(x)=(4x+3)b(x)-2,[/MATH]
with which by the rest theorem [MATH]p\left( \dfrac{3}{2}\right) =7[/MATH] and [MATH]p\left( -\dfrac{3}{4}\right) =-2[/MATH].

If [MATH]c(x)[/MATH] is the quotient of the division of [MATH]p(x)[/MATH] between [MATH]8x^2-6x-9[/MATH], that is, between [MATH](2x-3)(4x+3)[/MATH], again by the division algorithm, it is [MATH]p(x) = (2x-3)(4x + 3)c(x)+r(x)[/MATH], where [MATH]r(x)[/MATH] is the rest polynomial, whose degree is less than the degree of [MATH]8x^2-6x-9[/MATH], then [MATH]r(x)[/MATH] is of the form [MATH]mx+n[/MATH] for certain real constants [MATH]m[/MATH] and [MATH]n[/MATH].

Thus,

[MATH]7=p\left( \dfrac{3}{2}\right) =\dfrac{3m}{2}+n[/MATH]
and

[MATH]-2=p\left( -\dfrac{3}{4}\right) =-\dfrac{3m}{4}+n,[/MATH]
then [MATH]m=4[/MATH] and [MATH]n=1[/MATH] and, therefore, the polynomial rest requested in the statement is [MATH]r(x)=4x+1[/MATH].
 
You must log in or register to reply here.
Top