Question

If the polynomial px3+4x2+3x-4 and x3-4x+p are divided by (x-3), then the remainder in each case is the same, show that p= - 1

Answer 1

Let, f(x)=px^3+4x^2+3x-4

g(x)=x^3-4x+p

f(x) and g(x) are divided by (x-3).

same remainder,

f(3)=g(3)

p(3)^3+4(3)^2+3x3-4=(3)^3-4x3-p

27p+36+9-4=27-12+p

26p+41=15

26p=15-41

26p=-26

p=-26/26

p=-1

Answer 2

It has been shown that p= -1.

Given:

• Two polynomials.
• $p {x}^{3} + 4 {x}^{2} + 3x - 4$ and ${x}^{3} - 4x + p$
• Divided by (x-3)
• Remainder is same for both.

To find:

• Show that p=-1.

Solution:

Theorem to be used:

Remainder Theorem: If polynomial p(x) is divided by (x-a) then, remainder is given by p(a).

Step 1:

Find remainder for the first polynomial.

Let $f(x) = p {x}^{3} + 4 {x}^{2} + 3x - 4$

Put x= 3 in f(x).

$f(3) = p {(3)}^{3} + 4 {(3)}^{2} + 3(3) - 4$

or

$f(3) = 27p + 36 +9 - 4$

or

$\bf f(3) = 27p + 41...eq1$

Step 2:

Find remainder for second polynomial.

Let $g(x) = {x}^{3} - 4x + p$

Put x= 3 in g(x).

$g(3) = {(3)}^{3} - 4(3) + p$

or

$\bf g(3) =p + 15..eq2$

Step 3:

Equate both equations.

ATQ

$f(3) = g(3)$

or

$27p + 41 = p + 15$

or

$26p = - 26$

or

$\bf p = - 1$

Thus,

It has been shown that p= -1.

Learn more:

1) find the remainder when x⁴+ x³-2x²+ x + 1 is divided by x-1

https://brainly.in/question/9333959

2) Find the value of 'a' for which 2x3+ax2+11x+a+3 is exactly divisible by 2x-1

Please show the whole solution also because...

https://brainly.in/question/1194454