Question

What must be added to 3x³+x²-22x+9 so that the result is exactly divisible by 3x²+7x-6?

Answer 1

let f(x ) added 3x² + x² -22x +9
then result will be 3x² +7x -6

now,
f(x ) +3x2 +x² -22x +9 = 3x² +7x -6

f(x ) = 3x² +7x -6 -(3x² +x² -22x +9)
= (3 -3)x² +(0-1)x² + (7+22)x +(-6-9)
= -x² +29x -15

Answer 2

Answer:

∴ 2x + 3 is added to it.

Step-by-step explanation:

Let p(x) = 3x²+ x² - 22x + 9 and q(x) = 3x² + 7x - 6. When p(x) is divided by q(x), the remainder is linear expression in x.

So, let r(x) = ax + b is added to p(x) so that p(x) + r(x) is divisible by q(x).

Let, f(x) = p(x) + r(x). Then,

f(x) = 3(x)³ + x² - 22x + 9 + ax + b

= 3x³ + x² +x(a - 22) + b + 9

We have,

q(x) = 3x² + 7x - 6 = 3x² + 9x - 2x - 6 = 3x (x + 3) - 2 ( x + 3) = (3x - 2) (x + 3)

Clearly, q(x) is divisible by ( x + 3) and (3x - 2), i. e., (x + 3) and ( 3x - 2) are factors of q(x).

Therfore, f(x) will be divisible by q(x), if (x+3) and (3x-2) arr factors of f(x).

i.e. f(-3) = 0 and f(2/3) = 0

⟹ 3 x (-3)3 + (-3)2 + (-3) (a–22) b +9=0 [∵ f(-3) = 0]

⟹ -81 + 9 – 3(a – 22) + b+ 9 = 0

⟹ -81 + 9- 3a + 66 + b + 9 = 0

⟹ 3- 3a + b = 0

⟹ 3a – b = 3 .... (1)

And 3 x (2/3)³ + (2/3)² + 2/3(a-22)+b+9=0 [∵ f(2/3)=0]

⟹ 3 x 8/27 + 4/9 + 2a/3 - 44/3 + b + 9 =0

⟹ 8/9 + 4/9 + 2a/3 - 44/3 + b + 9 = 0

⟹ (8+4+6a-132+9b+81)/9=0

⟹ 6a + 9b - 39 = 0

⟹ 2a + 3b = 13

Multiplying equation (1) by 3 and adding equation (2), we get

11a = 22 ⟹ a = 2

Putting a = 2 in 2a + 3b = 13, we get

⟹ 2 x 2 + 3b = 13

⟹ 3b = 13 – 4

⟹ 3b = 9

⟹ b = 9/3

⟹ b = 3

∴ r(x) = ax + b = 2x + 3

Hence, 3x + x2 -22x +9 will be exactly divisible by 3x² + 7x - 6, if 2x+3 is added to it.