Question

Find the cubic polynomial whose
zeroes are -1, 1, 2​

Answer 1

Answer:

${x}^{3} - 2 {x}^{2} - x + 2$

Step-by-step explanation:

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Answer 2

Answer:

Given three zeroes are -1, 1, 2

Given three zeroes are -1, 1, 2p(X)=K[x^3-x^2(a+b+c)+X(ab+BC+ac-abc]

Given three zeroes are -1, 1, 2p(X)=K[x^3-x^2(a+b+c)+X(ab+BC+ac-abc]k[x^3-x^2(-1+1+2)+X(-1×2+1×2+2×-1]-(-1×1×2]

Given three zeroes are -1, 1, 2p(X)=K[x^3-x^2(a+b+c)+X(ab+BC+ac-abc]k[x^3-x^2(-1+1+2)+X(-1×2+1×2+2×-1]-(-1×1×2]k[x^3-x^2(2)+X(-2+2-2)-(-2)]

Given three zeroes are -1, 1, 2p(X)=K[x^3-x^2(a+b+c)+X(ab+BC+ac-abc]k[x^3-x^2(-1+1+2)+X(-1×2+1×2+2×-1]-(-1×1×2]k[x^3-x^2(2)+X(-2+2-2)-(-2)]k[x^3-2x^2+2x+2]

Given three zeroes are -1, 1, 2p(X)=K[x^3-x^2(a+b+c)+X(ab+BC+ac-abc]k[x^3-x^2(-1+1+2)+X(-1×2+1×2+2×-1]-(-1×1×2]k[x^3-x^2(2)+X(-2+2-2)-(-2)]k[x^3-2x^2+2x+2]let k=1

Given three zeroes are -1, 1, 2p(X)=K[x^3-x^2(a+b+c)+X(ab+BC+ac-abc]k[x^3-x^2(-1+1+2)+X(-1×2+1×2+2×-1]-(-1×1×2]k[x^3-x^2(2)+X(-2+2-2)-(-2)]k[x^3-2x^2+2x+2]let k=1x^3-2x^2+2x+2

Given three zeroes are -1, 1, 2p(X)=K[x^3-x^2(a+b+c)+X(ab+BC+ac-abc]k[x^3-x^2(-1+1+2)+X(-1×2+1×2+2×-1]-(-1×1×2]k[x^3-x^2(2)+X(-2+2-2)-(-2)]k[x^3-2x^2+2x+2]let k=1x^3-2x^2+2x+2.........................................................