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Let p(x) = x³ + 10x² + px + q

As (x – 1) and (x + 2) are factors of p(x) hence each will divide p(x) leaving remainder as 0

Let us use the remainder theorem which states that if (x – a) divides a quadratic polynomial p(x) then p(a) = 0 so we have p(1) = 0 and p(–2) = 0

⇒ p(1) = 0

⇒ 1³ + 10(1)² + p(1) + q = 0

⇒ 1 + 10 + p + q = 0

⇒ p + q = –11

⇒ q = –11 – p …(i)

⇒ p(–2) = 0

⇒ (–2)³ + 10(–2)² + p(–2) + q = 0

⇒ –8 + 40 – 2p + q = 0

⇒ 32 – 2p + q = 0

Substituting value of q from (i)

⇒ 32 – 2p – 11 – p = 0

⇒ 21 – 3p = 0

⇒ 3p = 21

⇒ p = 7

Substitute p in (i)

⇒ q = – 11 – 7

⇒ q = –18

Hence value of p and q are 7 and –18 respectively