Question

find the value of a for which pol polynomial 3 x cube + 14 x square + 9 X + a is divisible by 3X + 5


pls give solution​

Answer 1

Answer:

a=70

Step-by-step explanation:

$p(x) = {3x}^{3} + 14 {x}^{2} + 9x + a \\ p(x) = - 5 \\ p( - 5) = 3 \times ( { - 5}^{3} ) + 14 \times ( - {5}^{2} ) + 9 \times ( - 5) + a \\ p( - 5) = 3 \times - 125 + 14 \times 25 - 45 + a \\ p( - 5) = - 375 + 350 - 45 + a \\ p( - 5) = - 70 + a \\ butp( - 5) = 0 \\ - 70 + a = 0 \\ = a = 70$

Answer 2

Answer:

a=-10

Step-by-step explanation:

The given polynomial is

$P(x)=3x^3+14x^2+9x+a$

We need to find the value of a for which P(x) is divisible by (3x+5).

If a polynomial P(x) is completely divisible by (ax+b), then -b/a is a zero of P(x) and $P(\frac{-b}{a})=0$

Divisor is 3x+5, it means a=3 and b=5. P(x) is divisible by (3x+5), then

$P(\frac{-5}{3})=0$

$3(\frac{-5}{3})^3+14(\frac{-5}{3})^2+9(\frac{-5}{3})+a=0$

$3(\frac{-125}{27})+14(\frac{25}{9})-\frac{45}{3}+a=0$

$\frac{-125}{9}+\frac{350}{9}-\frac{45}{3}+a=0$

On further simplification, we get

$\frac{-125+350-135}{9}+a=0$

$\frac{90}{9}+a=0$

$10+a=0$

$a=-10$

Therefore, the value of a is -10.