If ax^3+9x^2+4x-10 when divided by x-3 leaves the reminder 5,then a=​

Let f(x) = ax³ + 9x² + 4x – 10

g(x) = 0⇒x - 3 = 0

⇒ x = 0 + 3

⇒ x = 3

On dividing f(x) by x - 3, it leaves a remainder 5.

Now keeping, f(3) = 5

⇒a(3)³ + 9(3)² + 4(3) - 10 = 5

⇒ a × 27 + 9 × 9 + 4 × 3 - 10 = 5

⇒ 27a + 81 + 12 - 10 = 5

⇒ 81 + 12 - 10 - 5 = 27a

⇒ 81 + 12 - 15 = 27a

⇒ 93 - 15 = 27a

⇒ 78 = 27a

⇒ a = 27/78

⇒ a = 0.3461

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

step-by-step explanation: