# Answer to Question #10521 in Algebra for Manthan khorwal

Question #10521

If (x-1/2) and (x-22) are the factors of the polynomials px2+5x+r,prove that p=r.

Expert's answer

We get:

px2+5x+r=k(x-1/2)(x-22)=kx2-23k/2+k/11 (k is some, nonzero number)

so p=k and r=k/11 p!=r

Obviously there is a mistake in condition, so the factors got to be (x-1/22) and (x-22), and

then:

px2+5x+r=k(x-1/22)(x-22)=kx2-485k/2+k

so p=k and r=k, and p=r.

px2+5x+r=k(x-1/2)(x-22)=kx2-23k/2+k/11 (k is some, nonzero number)

so p=k and r=k/11 p!=r

Obviously there is a mistake in condition, so the factors got to be (x-1/22) and (x-22), and

then:

px2+5x+r=k(x-1/22)(x-22)=kx2-485k/2+k

so p=k and r=k, and p=r.

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