Question

Which one of the following are zeroes of polynomial
p(x) = x2 – 2x - 8​

Answer 1

Answer:

x = 4 and -2

sum of zeros = - coeff of x / coeff of x2

product of zeros =

constant term / coeff of x2

Step-by-step explanation:

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Answer 2

$\mathcal{\green{\underline{\blue{GIVEN:}}}}$

• We are given a polynomial x²-2x-8.

$\mathcal{\green{\underline{\blue{TO \:\: FIND:}}}}$

• We need to find the zeroes of the above polynomial.

$\mathcal{\green{\underline{\blue{SOLUTION:}}}}$

There are many ways to find the zeros, but I am showing you two methods.

MIDDLE - TERM SPLITTING.

x²-2x-8=0.

• Multiply the coefficient of x² and the constant term. Here, the coefficient of x² is 1 and the constant is -8. Product = 1×(-8) = -8.
• Now, we need to split the middle term, that is, -2, in such way that the product is -8 and the sum is -2. There are two possibilities:

⇒ -4×2= -8; -4+2 = -2.

⇒ -8×1= -8; -8+1 = -7.

• Here, the first possibility, is correct, because it satisfies the condition stated above. So, we proceed splitting the middle term as -4 and 2.

$\sf \to x^2-2x-8=0$

$\sf \to x^2-4x+2x-8=0$

$\to \sf x(x-4)+2(x-4)=0$

$\to \sf{\blue{(x-4)(x+2)=0}}$

Thus, we can write that:

$\sf x-4=0; \:\: x+2=0; \:\: i.e.,$

$\boxed{\sf{\blue{x=4; \:\: x=-2}}}$

4 and -2 are the zeros.

QUADRATIC FORMULA.

The quadratic formula goes as:

For a equation in the form ax²+bx+c; the formula for x is:

$\boxed{\bold{\large{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}}}$

In the equation x²-2x-8, a=1, b= -2, c= -8. Substitute these values in the formula.

$\to \bold{\large{x = \frac{ -(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{ 2(1) }}}$

$\to \bold{\large{x = \frac{ 2 \pm \sqrt{36}}{ 2 }}}$

$\to \bold{\large{x = \frac{ 2 \pm 6 }{ 2 }}}$

Therefore,

$\to \bold{\large{x=\frac{2+6}{2}; \frac{2-6}{2} }}$

$\to \boxed{\bold{\large{x=4, -2}}}$

$\huge\star\:\:{\orange{\underline{\pink{\mathbf{HOPE \:\: THIS \:\: HELPS \:\: :D}}}}}$