Question

3. The other two zeroes of the polynomial x³-8x²+19x-12 if it's one zero is x-1 are
3,4
-3,-3
-3,4
3,4​

Answer 1

Question:

The other two zeroes of the polynomial x³-8x²+19x-12, if it's one zero is 1, are

Answer:

3, 4

Explanation:

p(x) = x³ - 8x² + 19x - 12

If 1 is one of its zeroes, it means

p(1) = 0

According to factor theorem, if p(a) = 0, then (x - a) is its factor.

So, (x - 1) is a factor of p(x).

(For the following 3 steps, refer to the images also.)

STEP 1

We know that (x - 1) is a factor of p(x) = x³ - 8x² + 19x - 12.

This means if we divide p(x) by (x - 1), we would get its other factors.

On dividing (x³ - 8x² + 19x - 12) by (x - 1), we get (x² - 7x + 12) as quotient.

(See the first image.)

This means that (x² - 7x + 12) is its quadratic factor.

STEP 2

Now, factorise (x² - 7x + 12) by middle term spliting to obtain two linear factors.

x³ - 8x² + 19x - 12 = (x-1)[x² - 7x + 12]

(See the second image.)

We get,

x³ - 8x² + 19x - 12 = (x - 1)(x - 4)(x - 3)

This means other than (x - 1), the factors of p(x) are (x - 4) and (x - 3)

STEP 3

We know that, to find the zeroes of a polynomial, we need to equate the polynomial to zero.

x³ - 8x² + 19x - 12 = 0

(x - 1)(x - 4)(x - 3) = 0

So, for this polynomial to be zero, one of these brackets has to be equal to zero. If one bracket equals 0, then automatically the whole polynomial becomes 0 as 0 multiplied by any number is 0.

(See the third image.)

Hence, the other two zeroes are 3 and 4.