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Factor Theorem

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Question

Find the zeros of the polynomial p(x)=x^2+2√2-6

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Solution

p(x) = x² + 2√2x - 6

We can find the zeros by factorization [ by splitting the middle terms ]

x² + 2√2x - 6
x² + 3√2x - √2x - 6

x [ x + 3√2 ] - √2 [ x+ 3√2]

[ x - √2 ] [ x+ 3√2]

the zeros are = √2 and -3√2

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Similar questions

Q. P(x)=x^2 + 2(√2)x - 6 find the zero of the polynomial and verify between zero and coefficients

Q. Verify that:
(i) 4 is a zero of the polynomial p(x) = x − 4.
(ii) −3 is a zero of the polynomial p(x) = x − 3.
(iii)

- \frac{1}{2}

is a zero of the polynomial p(y) = 2y + 1.
(iv)

\frac{2}{5}

is a zero of the polynomial p(x) = 2 − 5x.
(v) 1 and 2 are the zeros of the polynomial p(x) = (x − 1)(x − 2).
(vi) 0 and 3 are the zeros of the polynomial p(x) = x² − 3x.
(vii) 2 and −3 are the zeros of the polynomial p(x) = x² + x − 6.

Q. Find the zeros of the following polynomial:

p (x) = x - 2

Q. The zeros of the polynomial p(x) = x² + x – 6 are
(a) 2, 3
(b) –2, 3
(c) 2, –3
(d) –2, –3

Zero of the polynomial $p (x) = x^{2} + x - 6$ are
(a) 2,3
(b) -2,3
(c) 2,-3
(d) -2,-3

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