Answer:
#CarryOnLearning
\begin{gathered}(x + 4)(x + 3) \\ = {x}^{2} + 4x + 3x + 12 \\ = {x}^{2} + 7x + 12\end{gathered}
(x+4)(x+3)
=x
2
+4x+3x+12
+7x+12
\begin{gathered}(2k - 4)(k + 5) \\ = 2 {k}^{2} + 10k - 4k - 20 \\ = 2 {k}^{2} - + 6k - 20\end{gathered}
(2k−4)(k+5)
=2k
+10k−4k−20
−+6k−20
\begin{gathered}(y + 3)(y - 3) \\ = {y}^{2} + 3y - 3y - 9 \\ = {y}^{2} - 9\end{gathered}
(y+3)(y−3)
=y
+3y−3y−9
−9
\begin{gathered}(2c - 2)(2c + 2) \\ = 4 {c}^{2} - 4\end{gathered}
(2c−2)(2c+2)
=4c
−4
\begin{gathered} {(p + 4)}^{2} \\ = {p}^{2} + 8p + 16\end{gathered}
(p+4)
=p
+8p+16
\begin{gathered}(d - 5)(d - 5) \\ = {d}^{2} - 10d + 25\end{gathered}
(d−5)(d−5)
=d
−10d+25
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Answers & Comments
Answer:
#CarryOnLearning
Answer:
\begin{gathered}(x + 4)(x + 3) \\ = {x}^{2} + 4x + 3x + 12 \\ = {x}^{2} + 7x + 12\end{gathered}
(x+4)(x+3)
=x
2
+4x+3x+12
=x
2
+7x+12
\begin{gathered}(2k - 4)(k + 5) \\ = 2 {k}^{2} + 10k - 4k - 20 \\ = 2 {k}^{2} - + 6k - 20\end{gathered}
(2k−4)(k+5)
=2k
2
+10k−4k−20
=2k
2
−+6k−20
\begin{gathered}(y + 3)(y - 3) \\ = {y}^{2} + 3y - 3y - 9 \\ = {y}^{2} - 9\end{gathered}
(y+3)(y−3)
=y
2
+3y−3y−9
=y
2
−9
\begin{gathered}(2c - 2)(2c + 2) \\ = 4 {c}^{2} - 4\end{gathered}
(2c−2)(2c+2)
=4c
2
−4
\begin{gathered} {(p + 4)}^{2} \\ = {p}^{2} + 8p + 16\end{gathered}
(p+4)
2
=p
2
+8p+16
\begin{gathered}(d - 5)(d - 5) \\ = {d}^{2} - 10d + 25\end{gathered}
(d−5)(d−5)
=d
2
−10d+25
#CarryOnLearning