Question

Factorise the following polynomials by
proper substitution:
$( {{p}^{2} + 4p) }^{2} + 21( {p }^{2} + 4p) + 98$

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Answer 1

Let's simplify step-by-step.

• (p2+4p)2+21(p2+4p)+98

Distribute:

=p4+8p3+16p2+(21)(p2)+(21)(4p)+98

=p4+8p3+16p2+21p2+84p+98

Combine Like Terms:

=p4+8p3+16p2+21p2+84p+98

=(p4)+(8p3)+(16p2+21p2)+(84p)+(98)

=p4+8p3+37p2+84p+98

Answer:

=p4+8p3+37p2+84p+98[tex]

Answer 2

Answer:

$(p^2+4p)^2+21(p^2+4p)+98$

$p^2(p+4)^2+21p(p+4)+98$

$p^2(p+4)^2+7p(p+4)+14p(p+4)+98$

$p(p+4)[p(p+4)+7]+14[p(p+4)+7]$

$[p(p+4)+7][p(p+4)+14]$

$(p^2+4p+7)(p^2+4p+14)$