# Polynomials

## Short Questions

If a factor of 2x3 - 6x2 - 5m - 2 is x - 2, find the value of m.

Solution,
Let P(x) = 2x3 - 6x2 - 5m - 2
And d(x) = x - 2

Comparing d(x) with x - a, we get, a = 2.

Here, d(x) is factor of P(x).
Then, by the converse of factor theorem,
P(a) = 0
or, P(2) = 0
or, 2(2)3 - 6(2)2 - 5m - 2 = 0
or, 16 - 24 - 5m - 2 = 0
or, -10 = 5m
∴ m = -2

If (x + 2) is a factor of the polynomial 2x3 + 3x2 - mx + 4, find the value of m.

Solution,
Let P(x) = 2x3 + 3x2 - mx + 4
And d(x) = x + 2

Comparing d(x) with x - a, we get, a = -2.

Here, d(x) is factor of P(x).
Then, by the converse of factor theorem,
P(a) = 0 or, P(-2) = 0
or, 2(-2)3 + 3(-2)2 - m(-2) + 4 = 0
or, -16 + 12 + 2m + 4 = 0
or, 2m = 0
∴ m = 0

If 2x3 - 4x2 + kx + 10 is divided by (x + 2), then remainder os 4. Find the value of k using remainder theorem.

Given,
Polynomial p(x) = 2x3 - 42 + kx + 10
Divisor d(x) = x + 2
Remainder (R) = 4

Comparing d(x) with x - a, we get, a = -2

By remainder theorem,
R = p(a)
or, R = p(-2)
or, 4 = 2(-2)3 - 4(-2)2 + k(-2) + 10
or, 4 = -16 - 16 - 2k + 10
or, 20 + 16 = -2k + 10
or, 26 = -2k
∴ k = -13

If x - 3 is a factor of 2x3 - x2 + 10x - k then find the value of k.

Solution,
Let P(x) = 2x3 - x2 + 10x - k
And d(x) = x - 3

Comparing d(x) with x - a, we get, a = 3.

Here, d(x) is factor of P(x).
Then, by the converse of factor theorem,
P(a) = 0
or, P(3) = 0
or, 2(3)3 - 32 + 10(3) - k = 0
or, 54 - 9 + 30 - k = 0
or, 75 - k = 0
∴ k = 75

If x - 5 is a factor of x3 + px2 +4x + 5, find the value of p.

Solution,
Let P(x) = x3 + px2 + 4x + 5
And d(x) = x - 5

Comparing d(x) with x - a, we get, a = 5.

Here, d(x) is a factor of P(x).
Then, by the converse of factor theorem,
P(a) = 0
or, P(5) = 0
or, (5)3 + p(5)2 + 4(5) + 5 = 0
or, 125 + 25p + 20 + 5 = 0
or, 150 = -25p
∴ p = -6

Using Remainder theorem, find the remainder when 8x3 - 4x2 + 2x - 5 is divided by 2x - 1.

Given,
p(x) = 8x3 - 4x2 + 2x - 5
d(x) = 2x - 1

Comparing d(x) with ax - b, we get, a = 2 and b = 1.

By remainder theorem,
Remainder = p$\left(\frac{b}{a}\right)$
= p$\left(\frac{1}{2}\right)$
= 8$\left(\frac{1}{2}\right)^3$ - 4$\left(\frac{1}{2}\right)^2$ + 2$\left(\frac{1}{2}\right)$ - 5
= 1 - 1 + 1 - 5
= -4

If x - 5 is a factor of 2x3 - 7x + (p - 12), find the value of p.

Solution,
If p(x) = 2x3 - 7x + p - 12
And d(x) = x - 5

Comparing d(x) with x - a, we get a = 5

Here, d(x) is a factor of P(x)
Then, by the converse of factor theorem,
p(a) = 0
or, p(5) = 0
or, 2(5)3 - 7(5)p + p - 12 = 0
or, 250 - 35p + p - 12 = 0
or, 238 - 34p = 0
∴ p = 7

If x3 - 21x - 20 = (x + 1).q(x), find q(x) by using synthetic division method.

Given,
x3 - 21x - 20 = (x + 1).q(x)

Comparing with, p(x) = d(x).q(x), we get,
p(x) = x3 - 21x - 20
d(x) = x + 1

Comparing d(x) with x - a, we get, a = -1

Then, using synthetic division method, ∴ q(x) = x2 - x - 20

If x3 - 19x - 30 = (x + 2).q(x), find q(x) by using synthetic division method.

Given,
x3 - 19x - 30 = (x + 2).q(x)

Comparing with p(x) = d(x).q(x), we get,
p(x) = x3 - 19x - 30
d(x) = (x + 2)

Comparing d(x) with x - a, we get, a = -2

Then, using synthetic division method, ∴ q(x) = x2 - 2x - 15

State factor theorem. Use factor theorem to determine whether x + 3 is a factor of the polynomial x3 - 8x + 3.

Factor theorem: If P(x) is a polynomial of degree grater than 0 and P(a) = 0 then (x - a) is a factor of P(x). Conversely, if (x - a) is a factor of P(x), then P(a) = 0.

Solution,
p(x) = x3 - 8x + 3
d(x) = x + 3

Comparing d(x) with x - a, we get, a = -3

Now, to determine factor,
p(a) = p(-3)
= (-3)3 - 8(-3) + 3
= -27 + 24 + 3
= 0

Here, p(-3) = 0, that is remainder = 0. So, by factor theorem d(x) is a factor of p(x).

State the remainder theorem. When a polynomial x3 + 5x2 - 6x - 16 is divided by (x + 2), find the remainder by applying remainder theorem.

Remainder theorem: If P(x) is a polynomial of degree n and (x - a) is a divisor of P(x) then P(a) is a remainder, where the degree of quotient will be (n - 1).

Given,
p(x) = x3 + 5x2 - 6x - 16
d(x) = x + 2

Comparing d(x) with x - a, we get, a = -2

Now, By remainder theorem,
Remainder = p(a)
= p(-2)
= (-2)3 + 5(-2)2 - 6(-2) - 16
= -8 + 20 + 12 - 16
= 8

∴ Remainder = 8

If the polynomial x3 + 6x2 + kx + 10 is divided by (x + 2), the remainder is 4, find the value of k using the remainder theorem.

Given,
p(x) = x3 + 6x2 + kx + 10
d(x) = x + 2
Remainder (R) = 4

Comparing d(x) with x - a, we get, a = -2

Using remainder theorem,
R = p(a) or, 4 = p(-2)
or, 4 = (-2)3 + 6(2)2 - 2k + 10
or, 12 = 24 - 2k + 10
or, -22 = -2k
∴ k = 11

If (x + 2) is a factor of x3 - 19x - p, find the value of p.

Solution,
p(x) = x3 - 19x - p
d(x) = x + 2

Comparing d(x) with x - a, we get, a = -2

Here, d(x) is a factor of p(x).
Then, by the converse of factor theorem,
p(a) = 0
or, p(-2) = 0
or, (-2)3 - 19(-2) - p = 0
or, p = -8 + 38
∴ p = 30

If (x - m) is a factor of x3 - mx2 - 4x + m + 12, calculate the value of m.

Solution,
p(x) = x3 - mx2 - 4x + m + 12
d(x) = x - m

Comparing d(x) with x - a, we get, a = m.

Here, d(x) is factor of p(x).
Then, by the converse of factor theorem,
p(a) = 0
or, p(m) = 0
or, m3 - m3 - 4m + m + 12
or, -3m = -12
∴ m = 4

Define factor theorem. If x - 2 is a factor of x3 - kx2 + x - 6 = 0, find the value of 'k'.

Factor theorem: If P(x) is a polynomial of degree grater than 0 and P(a) = 0 then (x - a) is a factor of P(x). Conversely, if (x - a) is a factor of P(x), then P(a) = 0.

Solution,
p(x) = x3 - kx2 + x - 6 = 0
d(x) = x - 2

Comparing d(x) with x - a, we get, a = 2

Here, d(x) is a factor of p(x).
Then, by the converse of factor theorem,
p(a) = 0
or, p(2) = 0
or, (2)3 - k(2)2 + 2 - 6 = 0
or, 8 - 4k - 4 = 0
or, 4 = 4k
∴ k = 1

State factor theorem. Use factor theorem to determine whether x + 4 is a factor of the polynomial x3 - 15x + 4.

Factor theorem: If P(x) is a polynomial of degree grater than 0 and P(a) = 0 then (x - a) is a factor of P(x). Conversely, if (x - a) is a factor of P(x), then P(a) = 0.

Solution,
p(x) = x3 - 15x + 4
d(x) = x + 4

Comparing d(x) with x - a, we get, a = -4

Now,
p(a) = p(-4)
= (-4)3 - 15(-4) + 4
= -64 + 60 + 4
= 0

Here, p(a) = remainder = 0. So, using factor theorem d(x) is a factor of p(x).

State remainder theorem. Use it to find the remainder when the polynomial x6 - 1 is divided by x + 1.

Remainder theorem: If P(x) is a polynomial of degree n and (x - a) is a divisor of P(x) then P(a) is a remainder, where the degree of quotient will be (n - 1).

Solution,
p(x) = x6 - 1
d(x) = x + 1

Comparing d(x) with x - a, we get, a = -1

Now, using remainder theorem,
Remainder = p(a)
= p(-1)
= (-1)6 - 1
= (-1)2.(-1)2.(-1)2 - 1
= 1.1.1 - 1

∴ Remainder = 0

The polynomial f(x) = x3 - (P - 2)x2 - Px + 28 leaves a remainder 10 when divided by x + 3. Find the value of P, using the remainder theorem.

Given,
f(x) = x3 - (P - 2)x2 - Px + 28
Remainder(R) = 10
d(x) = x + 3

Comparing d(x) with x - a, we get, a = -3,

Using remainder theorem,
R = f(a)
or, 10 = f(-3)
or, 10 = (-3)3 - (P - 2)(-3)2 - Px + 28
or, 10 = -27 -(P - 2)9 + 3P + 28
or, 37 = -9P + 18 + 3P + 28
or, 9 = 18 - 6P
or, -9 = -6P
∴ P = $\frac{3}{2}$