Question

GIVEN THAT (X + 2) AND (X + 3) ARE THE FACTORS OF 2X^3 + AX^2 + 7X - B. DETERMINE THE VALUES OF A AND B?
(ANS - 9 AND 6) IS IT CORRECT PLEASE HELP ME!

Answer 1

EXPLANATION.

⇒ (x + 2) and (x + 3) are the factors.

⇒ 2x³ + ax² + 7x - b.

As we know that,

⇒ (x + 2) is a factors of equation.

⇒ x + 2 = 0.

⇒ x = - 2.

Put the value of x = - 2 in the equation, we get.

⇒ 2(-2)³ + a(-2)² + 7(-2) - b = 0.

⇒ - 16 + 4a - 14 - b = 0.

⇒ 4a - b - 30 = 0.

⇒ 4a - 30 = b. - - - - - (1).

⇒ (x + 3) is a factors of the equation.

⇒ x + 3 = 0.

⇒ x = - 3.

Put the value of x = - 3 in the equation, we get.

⇒ 2(-3)³ + a(-3)² + 7(-3) - b = 0.

⇒ - 54 + 9a - 21 - b = 0.

⇒ 9a - b - 75 = 0.

⇒ 9a - 75 = b. - - - - - (2).

From equation (1) and (2), we get.

We can write equation as,

⇒ 4a - 30 = 9a - 75.

⇒ - 30 + 75 = 9a - 4a.

⇒ 45 = 5a.

⇒ a = 9.

Put the values of a = 9 in the equation, we get.

⇒ 4a - 30 = b.

⇒ b = 4(9) - 30.

⇒ b = 36 - 30.

⇒ b = 6.

Values of a = 9 and b = 6.

Answer 2

Answer:

Yes, it is correct.

Step-by-step explanation:

Given, the polynomial

$\rm \: 2 {x}^{3} + A {x}^{2} + 7x - B$

and (x+2) and (x+3) are factors of that polynomial.

Let,

$\rm \: f(x) = 2 {x}^{3} + A {x}^{2} + 7x - B \\ \\ \\ \rm \: since \: (x + 2 )\: \: and \: \: (x + 3 )\: are \\ \rm \: the \: factors \: ,if \: we \: put \: the \: value \\ \rm \: of \: \: \pink{ x = - 2} \: \: and \: \: \pink{ x = - 3} \: in \: the \: \\ \rm \: polynomial \: then \: f(x) = 0 \\ \\ \therefore \rm \: f( - 2) = 2 \times {( - 2)}^{3} + A {( - 2)}^{2} + 7 \times ( - 2) - B \\ \\ \rm \implies \: - 16 + 4A - 14 - B = 0 \\ \\ \green{\rm \implies \:4A - B - 30 = 0 \: \: \: \: - - - - - (1)}\\ \\ \\ \bf \: again \: \\ \rm \: f( - 3) = 2 \times {( - 3)}^{3} + A {( - 3)}^{2} + 7 \times ( - 3) - B \\ \\ \rm \implies \: - 54 + 9A - 21 - B = 0 \\ \\ \red{ \rm \implies \:9A - B - 75 = 0 \: \: \: \: - - - - (2)} \\ \\ \sf \: eqn.(2) - eqn.(1) \implies \\ \\ \orange{ {\boxed{\begin{array}{cc}\rm \: 9A - B - 75 = 0 \\ \rm \: 4A - B - 30 = 0 \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \rm \: 5A - 45 = 0 \\ = > A = \frac{45}{5} = 9 \\ \\ \therefore \: A = 9 \end{array}}}}$

Put the value of A in eqn.(1)

$4 \times 9 - B - 30 = 0 \\ \\ = > B = 36 - 30 \\ \\ = > B = 6$

$\rm \therefore \: \: A = 9 \: \: \: \: and \: \: \: \: B = 6$