Question

Factorise the expression 5x^3+9x^2-96x-80 into three linear factories ,given (x+5) is a factor.

Answer 1

Divide the given polynomial with (x+5), u will get the quotient. factorise the quotient, and u get the other 2 factors.

Answer 2

how complete please ?

Answer 3

Wait.

Answer 4

We have
\[5x^3+9x^2-96x-80\]
It's given that x+5 is a factor, either we can divide the expression by x+5 to find a quadratic equation which will give us the two other factors or we could create x+5 factor in the expression and take it out to find our quadratic.
I'll show you the second one.

Answer 5

5x^(3)+9x^(2)-96x-80

(x-4)(x+5)(x+(4)/(5))

Answer 6

use synthetic division

-5 | 5 9 -96 -80
-25 80 80
-----------------
5 -16 -16 0

--> 5x^2 -16x -16

Answer 7

\[5x^3+9x^2-96x-80\]
Let's create x+5
we can write first term as
\[5x^3=5x^2(x+5)-25x^2\]
so
\[5x^2(x+5)-25x^2+9x^2-96x-80\]
or
\[5x^2(x+5)-16x^2-96x-80\]
Writing second term as
\[-16x^2=-16x(x+5)+80x \]
now we have
\[5x^2(x+5)-16x(x+5)+80x-96x-80\]
or
\[5x^2(x+5)-16x(x+5)-16x-80=>5x^2(x+5)-16x(x+5)-16(x+5)\]
so we get
\[(x+5)(5x^2-16x-16)\]
now let's factor
\[(5x^2-16x-16\]
we need to find factors of -16*5=-80 such that their sum is -16
-20 and 4 satisfy this so
\[(5x^2-20x+4x-16=>5x(x-4)+4(x-4)\]
we get
\[(x-4)(5x+4)\]
so (x-4) and (5x+4) are the two other factors

Answer 8

you can try by using long division..
or try any value..
subtitube any number in x the it will be the factor of that equation ..

Answer 9

thank you guys